暑期培訓第一周組隊賽
阿新 • • 發佈：2017-08-16
ger TTT 排序 ini 最大和 style alpha using cross
You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness‘‘, from ``most sorted‘‘ to ``least sorted‘‘. All the strings are of the same length.
Input The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.
Output Output the list of input strings, arranged from ``most sorted‘‘ to ``least sorted‘‘. Since two strings can be equally sorted, then output them according to the orginal order.
Sample Input
Sample Output
Garfield is not able to estimate the minimal steps to achieve the aim. Can you tell him?
Input For each line, there are three intergers Y, B, R(1<=Y,B,R<=1000),indicate the number refered above.
Output For each case, tell Garfield the minimal steps to complete the assignment. If not, output the symbol “):”.
Sample Input
Sample Output
是否AC的規則例如以下：
1. zoj能AC。
2. 若字符串形式為xzojx，則也能AC，當中x能夠是N個‘o‘ 或者為空。
3. 若azbjc 能AC，則azbojac也能AC，當中a,b,c為N個‘o‘或者為空；
Input 輸入包括多組測試用例。每行有一個僅僅包括‘z‘,‘o‘,‘j‘三種字符的字符串，字符串長度小於等於1000；
Output 對於給定的字符串。假設能AC則請輸出字符串“Accepted”，否則請輸出“Wrong Answer”。
Sample Output
Zty: I want to calculate N!......
WhereIsHeroFrom: So easy! How big N is ?
Zty: 1 <=N <=1000000000000000000000000000000000000000000000…
WhereIsHeroFrom: Oh! You must be crazy! Are you Fa Shao?
Zty: No. I haven‘s finished my saying. I just said I want to calculate N! mod 2009
Hint : 0! = 1, N! = N*(N-1)!
Input Each line will contain one integer N(0 <= N<=10^9). Process to end of file.
Output For each case, output N! mod 2009
Sample Input
Sample Output
Input The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.
Output The output contain n lines, each line output the number of result you can get .
Sample Input
Sample Output
Input The input contains four parts: 1) a dictionary, which consists of at least one and at most 100 words, one per line; 2) a line containing XXXXXX, which signals the end of the dictionary; 3) one or more scrambled ‘words‘ that you must unscramble, each on a line by itself; and 4) another line containing XXXXXX, which signals the end of the file. All words, including both dictionary words and scrambled words, consist only of lowercase English letters and will be at least one and at most six characters long. (Note that the sentinel XXXXXX contains uppercase X‘s.) The dictionary is not necessarily in sorted order, but each word in the dictionary is unique.
Output For each scrambled word in the input, output an alphabetical list of all dictionary words that can be formed by rearranging the letters in the scrambled word. Each word in this list must appear on a line by itself. If the list is empty (because no dictionary words can be formed), output the line "NOT A VALID WORD" instead. In either case, output a line containing six asterisks to signal the end of the list.
Sample Input
Sample Output
Sample Output
Sample Output
****-/*----/----*/****-/****-/*----/---**/*----/****-/*----/-****/***--/****-/*----/----*/**---/-****/**---/**---/***--/--***/****-/ He was so anxious that he asked for help in the Internet and after one day a girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:
1.First translate the morse code to a number string:4194418141634192622374
2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS
3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ
So ,we can get OTOEOIOUYVL
4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI
5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO
I guess you might worship Pianyi angel as me,so let‘s Orz her.
Now,the task is translate the number strings.
Input A number string each line(length <= 1000). I ensure all input are legal.
Output An upper alphabet string.
Sample Input
Sample Output
有例如以下所看到的的數塔。要求從頂層走究竟層，若每一步僅僅能走到相鄰的結點。則經過的結點的數字之和最大是多少？
已經告訴你了。這是個DP的題目，你能AC嗎?
Sample Output
DNA Sorting
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 3
Problem Description One measure of ``unsortedness‘‘ in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC‘‘, this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG‘‘ has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM‘‘ has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted).You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness‘‘, from ``most sorted‘‘ to ``least sorted‘‘. All the strings are of the same length.
Input The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.
Output Output the list of input strings, arranged from ``most sorted‘‘ to ``least sorted‘‘. Since two strings can be equally sorted, then output them according to the orginal order.
Sample Input
10 6 AACATGAAGG TTTTGGCCAA TTTGGCCAAA GATCAGATTT CCCGGGGGGA ATCGATGCAT
Sample Output
CCCGGGGGGA AACATGAAGG GATCAGATTT ATCGATGCAT TTTTGGCCAA TTTGGCCAAA
////大意是以上面提到的方式排序，安排好序的字符串輸出
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct stu
{
char s[100];
int cnt;
}arr[100];
int cmp(struct stu a,struct stu b)
{
//if(a.t==b.t) return 0;
return a.cnt<b.cnt;//升序
}
int main()
{
int m,n;
int i,j,k,l;
int t;
scanf("%d",&t);
getchar();
while(t--)
{
scanf("%d%d",&m,&n);
getchar();
for(i=0;i<100;++i)
{
arr[i].cnt=0;
}
for(i=0;i<n;++i)
{
scanf("%s",arr[i].s);
}
for(i=0;i<n;i++)//題目中提到的方式 把得數存到一個數組裏
{
for(k=0;k<m;k++)
{
for(j=k;j<m;j++)
{
if(arr[i].s[k]>arr[i].s[j])
arr[i].cnt++;
}
}
}
sort(arr,arr+n,cmp);
for(i=0;i<n;++i)
{
printf("%s\n",arr[i].s);
}
}
return 0;
}
Change the ball
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 26 Accepted Submission(s) : 2
Problem Description Garfield has three piles of balls, each pile has unique color of following: yellow, blue, and red. Now we also know Garfield has Y yellow balls, B blue balls, and R red balls. But Garfield just wants to change all the balls to one color. When he puts two balls of different color togather, the balls then change their colors automatically into the rest color. For instance, when Garfield puts a red one and a yellow one togather, the two balls immediately owns blue color, the same to other situations. But the rule doesn’t work when the two balls have the same color.Garfield is not able to estimate the minimal steps to achieve the aim. Can you tell him?
Input For each line, there are three intergers Y, B, R(1<=Y,B,R<=1000),indicate the number refered above.
Output For each case, tell Garfield the minimal steps to complete the assignment. If not, output the symbol “):”.
Sample Input
1 2 3 1 2 2
Sample Output
): 2
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
int a[3];
while(scanf("%d%d%d",&a[0],&a[1],&a[2])!=EOF)
{
sort(a,a+3);
int step=0;
if((a[1]-a[0])%3==0)
step=a[1];
else if((a[2]-a[1])%3==0)
step=a[2];
else if((a[2]-a[0])%3==0)
step=a[2];
else {
printf("):\n");continue;
}
printf("%d\n",step);
}
return 0;
} ZOJ問題
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 44 Accepted Submission(s) : 4
Problem Description 對給定的字符串(僅僅包括‘z‘,‘o‘,‘j‘三種字符),推斷他能否AC。是否AC的規則例如以下：
1. zoj能AC。
2. 若字符串形式為xzojx，則也能AC，當中x能夠是N個‘o‘ 或者為空。
3. 若azbjc 能AC，則azbojac也能AC，當中a,b,c為N個‘o‘或者為空；
Input 輸入包括多組測試用例。每行有一個僅僅包括‘z‘,‘o‘,‘j‘三種字符的字符串，字符串長度小於等於1000；
Output 對於給定的字符串。假設能AC則請輸出字符串“Accepted”，否則請輸出“Wrong Answer”。
Sample Input
zoj ozojo ozoojoo oozoojoooo zooj ozojo oooozojo zojoooo
Sample Output
Accepted Accepted Accepted Accepted Accepted Accepted Wrong Answer Wrong Answer
//ZOJ問題
//1. zoj能AC。
//2. 若字符串形式為xzojx，則也能AC，當中x能夠是N個'o' 或者為空；
//3. 若azbjc 能AC。則azbojac也能AC。當中a,b,c為N個'o'或者為空；
//若azbjc 能AC，則azbojac也能AC
//這裏中間加了一個o。右邊加了a個o，找到了一個規律。就是：
//a*b==c即Aceeptd，a表示z之前o的個數，
//b表示zj之間o的個數。c表示j之後o個數
#include<stdio.h>
#include<string.h>
int main()
{
int a,b,c,i;//a表示z之前o的個數，
//b表示zj之間o的個數，c表示j之後o個數
char s[1010];
while(gets(s))
{
//a=b=c=0;
int len=strlen(s);
for(i=0;i<len;i++)
{
if(s[i]=='z')//找到z，它前面的都是o
a=i;
if(s[i]=='j')
b=i-a-1;//找到j，算出中間的o
}
c=len-a-b-2;
if(a==0)
{
if(b>0&&c==0)
{
printf("Accepted\n");
}
else printf("Wrong Answer\n");
}
else
{
if(a*b==c&&b>0) //b>0一定要加，可能會出現oozj
printf("Accepted\n");
else printf("Wrong Answer\n");
}
//printf("%d %d %d\n\n",a,b,c);
}
return 0;
}
/*
N！Again
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 24 Accepted Submission(s) : 10
Problem Description WhereIsHeroFrom: Zty, what are you doing ?Zty: I want to calculate N!......
WhereIsHeroFrom: So easy! How big N is ?
Zty: 1 <=N <=1000000000000000000000000000000000000000000000…
WhereIsHeroFrom: Oh! You must be crazy! Are you Fa Shao?
Zty: No. I haven‘s finished my saying. I just said I want to calculate N! mod 2009
Hint : 0! = 1, N! = N*(N-1)!
Input Each line will contain one integer N(0 <= N<=10^9). Process to end of file.
Output For each case, output N! mod 2009
Sample Input
4 5
Sample Output
24 120
//2009=7*7*41,41以後的數對2009求余都是0，離散中講過mod k。mod就是求余的意思
#include<stdio.h>
int main()
{
int i,n;
while(~scanf("%d",&n))
{
int s=1;
if(n>=41) printf("0\n");
else
{
for(i=2;i<=n;++i)
{
s=(s*i)%2009;
}
printf("%d\n",s);
}
}
return 0;
}
1sting
Time Limit : 5000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 36 Accepted Submission(s) : 10
Problem Description You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.Input The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.
Output The output contain n lines, each line output the number of result you can get .
Sample Input
3 1 11 11111
Sample Output
1 2 8
//做完fibonacci和大菲波數就能懂了
#include<stdio.h>
#include<string.h>
char s[210];
int num[210][1010];
int len;
int main()
{
int n;
int i,j;
memset(num,0,sizeof(num));
num[1][0]=1;
num[2][0]=2;
//int k,l;
//l=0;
for(i=3;i<210;i++)
{
//k=0;
for(j=0;j<1000;j++)
{
//l=k+num[i-1][j]+num[i-2][j];
//num[i][j]=l%10;
//k=l/10;
num[i][j]+=num[i-1][j]+num[i-2][j];//'+='把我坑慘了
if(num[i][j]>=10)
{
num[i][j]-=10;
num[i][j+1]+=1;
}
}
}
scanf("%d",&n);
getchar();
while(n--)
{
scanf("%s",s);
len=strlen(s);
for(i=1000-1;i>0&&num[len][i]==0;--i)//清零
;
for(;i>=0;i--)
printf("%d",num[len][i]);
printf("\n");
}
return 0;
}
Word Amalgamation
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 5 Accepted Submission(s) : 3
Problem Description In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find the letters that appear in the answer it is necessary to unscramble four words. Your task is to write a program that can unscramble words.Input The input contains four parts: 1) a dictionary, which consists of at least one and at most 100 words, one per line; 2) a line containing XXXXXX, which signals the end of the dictionary; 3) one or more scrambled ‘words‘ that you must unscramble, each on a line by itself; and 4) another line containing XXXXXX, which signals the end of the file. All words, including both dictionary words and scrambled words, consist only of lowercase English letters and will be at least one and at most six characters long. (Note that the sentinel XXXXXX contains uppercase X‘s.) The dictionary is not necessarily in sorted order, but each word in the dictionary is unique.
Output For each scrambled word in the input, output an alphabetical list of all dictionary words that can be formed by rearranging the letters in the scrambled word. Each word in this list must appear on a line by itself. If the list is empty (because no dictionary words can be formed), output the line "NOT A VALID WORD" instead. In either case, output a line containing six asterisks to signal the end of the list.
Sample Input
tarp given score refund only trap work earn course pepper part XXXXXX resco nfudre aptr sett oresuc XXXXXX
Sample Output
score ****** refund ****** part tarp trap ****** NOT A VALID WORD ****** course ******
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
struct node
{
char s[10];
char st[10];//已經排好序的字符串
};
node arr[103];
int cmp(node a,node b)
{
return strcmp(a.s,b.s)<=0;
}
int main()
{
char s[10];
int i,n=0;
while(scanf("%s",s),strcmp(s,"XXXXXX")!=0)
{
strcpy(arr[n].s,s);//這裏用得特別好，這樣就能夠把原來的字符串和如今的字符串分別存入
sort(s,s+strlen(s));
strcpy(arr[n].st,s);
n++;
}
sort(arr,arr+n,cmp);//依照字典順序將字符串排序
bool flag;
while(scanf("%s",s),strcmp(s,"XXXXXX")!=0)
{
sort(s,s+strlen(s));
flag=0;
for(i=0;i<n;i++)//全部的都比較一下
{
if(strcmp(arr[i].st,s)==0)//排序完後比較兩都排好序的字符串
{
flag=1;printf("%s\n",arr[i].s);
}
}
if(!flag)
printf("NOT A VALID WORD\n");
printf("******\n");//每次找完一個都要輸出這個
}
return 0;
}
kiki‘s game
Time Limit : 5000/1000ms (Java/Other) Memory Limit : 40000/10000K (Java/Other)
Total Submission(s) : 24 Accepted Submission(s) : 15
Problem Description Recently kiki has nothing to do. While she is bored, an idea appears in his mind, she just playes the checkerboard game.The size of the chesserboard is n*m.First of all, a coin is placed in the top right corner(1,m). Each time one people can move the coin into the left, the underneath or the left-underneath blank space.The person who can‘t make a move will lose the game. kiki plays it with ZZ.The game always starts with kiki. If both play perfectly, who will win the game?
Input
Input contains multiple test cases. Each line contains two integer n, m (0<n,m<=2000). the="" input="" is="" terminated="" when="" n="0" and="" m="0." <="" div="">
Output
If kiki wins the game printf "Wonderful!", else "What a pity!".
Sample Input
5 3 5 4 6 6 0 0
Sample Output
What a pity! Wonderful! Wonderful!
#include<stdio.h>
int main()
{
int m,n;
while(scanf("%d%d",&m,&n),m+n)
{
if(m%2==0||n%2==0) printf("Wonderful!\n");
else printf("What a pity!\n");
}
return 0;
}
/*
我們把Win的地方標記成W，Lose的地方標記成L
W W W W W W W W
L W L W L W L W
W W W W W W W W
L W L W L W L W
W W W W W W W W
L W L W L W L W
我們能夠找到規律，當m%2==0||n%2==0，就Win了
*/
疊筐
Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 52 Accepted Submission(s) : 11
Problem Description 須要的時候，就把一個個大小差一圈的筐疊上去，使得從上往下看時，邊筐花色交錯。這個工作如今要讓計算機來完畢。得看你的了。
Input
輸入是一個個的三元組，各自是，外筐尺寸n（n為滿足0<n<80的奇整數）。中心花色字符，外筐花色字符。後二者都為ascii可見字符；< div="">
Output
輸出疊在一起的筐圖案，中心花色與外筐花色字符從內層起交錯相疊，多筐相疊時，最外筐的角總是被打磨掉。疊筐與疊筐之間應有一行間隔。
Sample Input
11 B A 5 @ W
Sample Output
AAAAAAAAA ABBBBBBBBBA ABAAAAAAABA ABABBBBBABA ABABAAABABA ABABABABABA ABABAAABABA ABABBBBBABA ABAAAAAAABA ABBBBBBBBBA AAAAAAAAA @@@ @WWW@ @[email protected]@ @WWW@ @@@
#include<stdio.h>
char map[100][100];
int main()
{
int n,i,j,mark;
char a,b,t,flag=0;
while(~scanf("%d %c %c",&n,&a,&b))
{
if(flag) putchar('\n');
flag=1;
if((n-1)%4)//推斷哪個在最外面，這個是找規律
{
t=a;
a=b;
b=t;
}
t=a,mark=1;//t和mark的初始化
for(i=0;i<=n/2;i++)
{
t=a;
mark=1;
for(j=0;j<n;j++)
{
if(n!=1&&i==0&&(j==0||j==n-1))//註意n==1的時候
{
putchar(' ');
map[i][j]=' ';
continue;
}
map[i][j]=t;
putchar(t);
if(i>j||j>=(n-1)-i)//這也是找規律，什麽時候要變
{
if(mark)//輪流變化
{
t=b;
mark=0;
}
else
{
t=a;
mark=1;
}
}
}
putchar('\n');
}
for(i=n/2-1;i>=0;i--)
{
for(j=0;j<n;j++)
{
printf("%c",map[i][j]);
}
putchar('\n');
}
}
return 0;
}
I Love You Too
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 11 Accepted Submission(s) : 5
Problem Description This is a true story. A man showed his love to a girl,but the girl didn‘t replied clearly ,just gave him a Morse Code:****-/*----/----*/****-/****-/*----/---**/*----/****-/*----/-****/***--/****-/*----/----*/**---/-****/**---/**---/***--/--***/****-/ He was so anxious that he asked for help in the Internet and after one day a girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:
1.First translate the morse code to a number string:4194418141634192622374
2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS
3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ
So ,we can get OTOEOIOUYVL
4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI
5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO
I guess you might worship Pianyi angel as me,so let‘s Orz her.
Now,the task is translate the number strings.
Input A number string each line(length <= 1000). I ensure all input are legal.
Output An upper alphabet string.
Sample Input
4194418141634192622374 41944181416341926223
Sample Output
ILOVEYOUTOO VOYEUOOTIO
#include<stdio.h>
#include<string.h>
int i,len,j;
char s1[1010],s2[1010];
char s[11][5]={"_","ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
char str[]={"QWERTYUIOPASDFGHJKLZXCVBNM"};
char s3[1010],s4[1010];
char res[1010];
int main()
{
while(gets(s1))
{
len=strlen(s1);
for(i=0,j=0;i<len;i+=2)//數字轉換為字母
{
s2[j++]=s[s1[i]-'0'-1][s1[i+1]-'0'-1];
}
int k;
for(i=0;i<j;++i)//依照題目給的方式進行字符變化
{
for(k=0;k<26;++k)
if(str[k]==s2[i])
{
s1[i]='A'+k;break;
}
}
len=len/2+1;
for(i=0;i<len/2;++i)//把一個字符串分成兩
{
s3[i]=s1[i];
}
s3[i]='\0';
int t=i;
for(j=0;i<len-1;++i)
{
s4[j++]=s1[i];
}
s4[j]='\0';
int l1=strlen(s3),l2=strlen(s4);
int flag=1;
i=0,j=0;
char ch;
while(s4[i]!='\0')//輪流存到res數組裏
{
if(!flag)
{
flag=1;
ch=s4[i];
++i;
//if(s4[i]=='\0') continue;
}
else
{
flag=0;
ch=s3[i];
}
res[j++]=ch;
}
if(l1!=l2) res[j++]=s3[i];
/*
k=0;
for(i=0;i<j;++i,k+=2)
{
res[k]=s3[i];
res[k+1]=s4[i];
}
for(;i<t;++i)
{
res[k]=s3[i];
++k;
}
*/
for(i=j-1;i>=0;--i)
{
printf("%c",res[i]);
}
puts("");
}
return 0;
}數塔
Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 15 Accepted Submission(s) : 11
Problem Description 在講述DP算法的時候。一個經典的樣例就是數塔問題，它是這樣描寫敘述的：有例如以下所看到的的數塔。要求從頂層走究竟層，若每一步僅僅能走到相鄰的結點。則經過的結點的數字之和最大是多少？
已經告訴你了。這是個DP的題目，你能AC嗎?
Input
輸入數據首先包含一個整數C,表示測試實例的個數，每一個測試實例的第一行是一個整數N(1 <= N <= 100)，表示數塔的高度。接下來用N行數字表示數塔，當中第i行有個i個整數。且全部的整數均在區間[0,99]內。
Output
對於每一個測試實例，輸出可能得到的最大和，每一個實例的輸出占一行。
Sample Input
1 5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
#include<stdio.h>
#define Max(a,b) (a)>(b)?(a):(b)
int a[100][110];
int main()
{
int i,j;
int t,n,m;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=1;i<=n;++i)//把數存入
{
for(j=1;j<=i;++j)
scanf("%d",&a[i][j]);
}
for(i=n;i>1;--i)//從下往上加
{
for(j=i;j>1;--j)
{
a[i-1][j-1]+=Max(a[i][j],a[i][j-1]);//把以下兩個的較大的一個和上面那個數相加
}
}
printf("%d\n",a[1][1]);//一直加到第一個，輸出答案
}
return 0;
}暑期培訓第一周組隊賽