1035: UFO

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 14  解決: 9
[提交][狀態][討論版]

題目描述

long long ago, there's a poor guy zyyyyy. one day he met a Huge Hovering UFO(HHU) and asked aliens in the HHU to make him rich, the aliens agreed and they gave him N problems to solve, and promised him if he solve problem i, he will get a[i] yuan for reward. However the aliens are evil, in fact they can get b[i] bytes of the knowledge of human beings if zyyyyy solve problem i. if they their knowledge sum up to more than M, they will explode the world. zyyyyy is very smart, he got the most money without exploding the world and made the aliens very angry. So can you figure out how can he do that?

輸入

there are several test cases. every test case contains N + 1 lines:
line 1: N(1 <= N <= 3000), M(1 <= M <= 12000)
line 2..N+1: line i + 1 gives a[i] and b[i], (1 <= a[i] <= 400, 1 <= b[i] <= 100)
both N, M, a[i], b[i] are integers

輸出

for each test case, output the money zyyyyy eventually got

樣例輸入

4 6
4 1
6 2
12 3
7 2

樣例輸出

23

提示

來源

//B
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<map>
using namespace std;
const int maxn =  12010;
int f[maxn];
int a[3010];
int b[3010];
int main(){
    int n,m,i,j;
    while(~scanf("%d%d",&n,&m)){
        memset(f,0,sizeof(f));
        for(i=1;i<=n;i++){
            scanf("%d%d",&a[i],&b[i]);
        }
        for(i=1;i<=n;i++){
            for(j=m;j>=b[i];j--){
                f[j] = max(f[j-b[i]]+a[i],f[j]);
            }
        }
        printf("%d\n",f[m]);
    }
    return 0;
}
 

1036: HHU tower

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 10  解決: 3
[提交][狀態][討論版]

題目描述

there is a tower in the universe's center - Hohai University, people call it HHU tower. HHU tower has 3 pillars(柱子). there is N plates(盤子) through these pillars which are labeled from 1 to N. HHU tower has a strange feature that for every pillar, plates with smaller index are always above plates with bigger index at any time. for example, plate i must be above plate i + 1 if they are placed through the same pillar. Originally, all the plates are placed through pillar 1, the big brother of HHU ask zyyyyy to move one plate from one pillar to the other(must satisfy HHU tower's feature) each day until all the plates are finally placed through pillar 3. zyyyyy is poor so he want to keep this job as long as possible. However the big brother know that zyyyyy may play trick so he can do this job forever so he ask zyyyyy to make sure that the pillars' state must not be repeated. zyyyyy knows the big brother is watching him, so please help him find how many days can he do this job at most?

輸入

there are several test cases. each test case has one line with a integer N(1 <= N <= 10000)

輸出

for each N, output the most steps need. the result can be very large so mod 10000007.

樣例輸入

1

樣例輸出

2

提示

origin state: the only pillar is through pillar 1

day 1: move the only plate from pillar 1 to pillar 2

day 2: move the only plate from pillar 2 to pillar 3, you can not move to pillar 1 again because that will repeat the origin state.

來源

這個是快速冪。。。然而公式是如何推出來的呢。。。

小萌學姐：

你看啊，這個總共有三個柱子n個盤子嘛，要求每次盤子的擺放狀態都要不一樣啊，而且還要取最大嘛，那就是3^n次方嘍，結果要減一嘛，因為去掉最開始的那個狀態

原來如此。。。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <cstdlib>
#include <queue>
#include <stack>
using namespace std;
long long mod = 10000007;
long long queen(long long another){
    if(another==0){
        return 1;
    }else if(another==1){
        return 3;
    }
    long long res = 0;
    res = queen(another/2)%mod;
    res = (res * res)%mod;
    if(another%2){
        res = (res * 3 )%mod;
    }
    return res;
}
int main(){
    long long n;
    while (~scanf("%lld",&n)) {
        long long ans = queen(n);
        printf("%lld\n",(ans-1)%mod);
    }
    return 0;
}

1037: Calvin's Experiment

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 2  解決: 2
[提交][狀態][討論版]

題目描述

Calvin Neo is a acmer in HHU, but he is also a civil engineer. Civil Engineering is about moving bricks(搬磚), however it involves many scientific theories and experiments.
Calvin designed a experiment to evaluate the attribute of soil in a construction site(工地), the construction site can be viewed as a 2-dimension Cartesian Coordinate System(平面直角座標系), this experiment includes 2 steps:
1. Calvin choose some point and use CDT(a geological survey method) to test the attribute of soils at that point and use them to work out some assumptions.
2. Calvin choose n tested points from step 1 and get a polygon(多邊形). he start from point 1 and go along the polygon, visit every point once and finally back to point 1. while walking, Calvin use geological radar(地質雷達) to detect the attribute of soils beneath to prove his assumptions
however life in Calvin's construction site is hard, there are ways between only several tested points and the geological radar is expensive so that Calvin should find a polygon with the shortest length(周長).
Calvin is a acmer and he definitely knows the answer, but he is busy moving bricks now, so he ask you for help.

輸入

there are multiple test cases, every test case has multiple lines:
line 1: 2 integers n(less than 233), m(less than 500, for 70% cases, m less than 200)
following n lines: 2 integers x, y stand for the coordinate of points(start from 1), 0 <= (x, y) <= 1000
following m lines: 2 integers u, v means there is a path between u and v

輸出

for every test case print one line, which is the minimum length(周長)
print "impossible" in one line if there is no such polygon
hint: you should use "%.2f\n" to print

樣例輸入

9 6
0 0
0 1
0 2
1 0
1 1
1 2
2 0
2 1
2 2
1 3
1 7
7 9
3 9
2 5
4 5
1 0
1 1

樣例輸出

4.00
impossible

提示

for case 1, choose point 1, 2, 4, 5 as a polygon whose length is 4.00

來源

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn = 250;
const double killme = 1e7;
const double eps = 1e-8;
double min_loop;
double mp[maxn][maxn];
double dis[maxn][maxn];
pair<double, double > save[maxn];
vector<int> final[510];

int n,m;
void init(){
    int i,j;
    for(i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            if(i==j){
                mp[i][j] = 0.0f;
                dis[i][j] = 0.0f;
            }else{
                mp[i][j] = INF;
                dis[i][j] = INF;
            }
        }
    }
    for(i=1;i<=m;i++){
        final[i].clear();
    }
}
bool online(int x,int y,int x1,int y1,int x2,int y2){
    int min_x = min(x1,x2),max_x = max(x1,x2);
    int min_y = min(y1,y2),max_y = max(y1,y2);
    if(min_x>x||x>max_x||y<min_y||y>max_y){
        return false;
    }
    if((y1 - y2) *  (x - x1) == (y - y1) * (x1 - x2)){
        return true;
    }
    return false;
}
bool floyd(){
    min_loop = INF;
    int i,j,k;
    for(k=1;k<=n;k++){
        for(i=1;i<k;i++){
            for(j=i+1;j<k;j++){
                if(min_loop>mp[i][j]+dis[i][k]+dis[k][j]){
                    min_loop = mp[i][j] + dis[i][k]+dis[k][j];
                }
            }
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                mp[i][j] = min(mp[i][j],mp[i][k]+mp[k][j]);
            }
        }
    }
    if(min_loop>killme){
        return  false;
    }else{
        return true;
    }
 
}
bool compare(int a,int b){
    if(abs(save[a].first-save[b].second)==0){
        return save[a].second<save[b].second;
    }
    return save[a].first<save[b].first;
}
int main(){
    //cout<<online(18, 97, 99, 67, 18, 97)<<endl;
    int i,j;
    while (~scanf("%d%d",&n,&m)) {
        double a,b;
        for(i=1;i<=n;i++){
            scanf("%lf%lf",&a,&b);
            save[i] = make_pair(a, b);
        }
        init();
        int u,v;
        for(i=1;i<=m;i++){
            scanf("%d%d",&u,&v);
            for(j=1;j<=n;j++){
                if(online(save[j].first, save[j].second, save[u].first, save[u].second, save[v].first, save[v].second)){
                    final[i].push_back(j);
                }
            }
            sort(final[i].begin(), final[i].end(), compare);
            for(j=0;j < final[i].size()-1;j++){
                u = final[i][j];
                v = final[i][j+1];
                mp[u][v] = min(mp[u][v] , sqrt(pow(save[u].first-save[v].first,2)+pow(save[u].second-save[v].second,2)));
                mp[v][u] = mp[u][v];
                dis[u][v] = mp[u][v];
                dis[v][u] = mp[u][v];
            }
            
        }
        if(floyd()){
            printf("%.2lf\n",min_loop);
        }else{
            printf("impossible\n");
        }
    }
    return 0;
}

1038: The Smartest Guy in HHU

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 11  解決: 5
[提交][狀態][討論版]

題目描述

Song and zyyyyy are all very smart guys in HHU and one day they asked Calvin to tell who is smarter. However later that day Song came to Calvin Neo, gave him some money and asked him to hard point(欽定) her to be smarter. Because the money was so much that Calvin no longer need to moving bricks(搬磚) to feed himself so they made a transaction.
The next day, Calvin introduce a game to Song and zyyyyy. In this game, Song and zyyyyy take turns to multiply an integer p by integer from 2 to 9. In each round, Song always starts with p = 1, does her multiplication, then zyyyyy multiplies the number, then Song and so on. Before a game starts, Calvin decides an integer 1 < n < 4294967295 and the winner is who first reaches p >= n. Because both Song and zyyyyy are smart, so they all play perfectly.
Calvin has some n, please help him decide which of them can make Song win.

輸入

one line for one case with a integer n

輸出

"yes" if Song can win, "no" otherwise

樣例輸入

162
17
34012226

樣例輸出

yes
no
yes

提示

zyyyyy and sxm are not real people

來源

小萌學姐：

哎呀給你一個數字，這個數字是目標，那上一個操作的人的最優解肯定是拿9乘以原先那個數字對不對，儘量把數字拉高嘛

那上上個人看自己快輸了肯定要儘量拉低數字嘛，所以肯定選最小的2來乘以原先的數

那麼把給我的數用18除到[0,18]的時候看一看到底是在[0,9]還是(9,18]就好了啊

嗯。。。。博弈論真是神奇啊

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <cstdlib>
#include <queue>
#include <stack>
using namespace std;
const double eps = 1e-8;
bool flag;
int main(){
    double n;
    bool flag;
    while(~scanf("%lf",&n)){
        flag = true;
        while (n>18.0) {
            n /= 18.0;
        }
        if(n<9.0||abs(n-9.0)<eps){
            flag = true;
        }else{
            flag = false;
        }
        if(flag){
            printf("yes\n");
        }else{
            printf("no\n");
        }
    }
    return 0;
}

1039: Zyyyyy和Song的閏年問題

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 13  解決: 8
[提交][狀態][討論版]

題目描述

    某天，Zyyyyy和Song一起學習有關閏年的知識，他們覺得我們平時所說的閏年是有問題的，為什麼呢？眾所周知，閏年(Leap Year)是為了彌補因人為曆法規定造成的年度天數與地球實際公轉週期的時間差而設立的，補上時間差的年份為閏年。四季構成的一年,就是“迴歸年”,也稱“太陽年”，即太陽中心從春分點再到春分點所經歷的時間，1迴歸年=365.24219879日。而我們習慣性的每四年一閏年。
     Zyyyyy覺得“每四年一閏年”是有問題，按照一年365天計算的話，如果n年之後，多出來的日子滿一天的話，這一年就該算作閏年。而Song覺得Zyyyyy才有問題，兩人爭執不休，於是找到了崇尚真理的你。
     在假設從公元0年開始的情況下，他們給定一個年份y，問你這個年份從真理的角度（即大佬Zyyyyy的角度）講，是不是閏年。

輸入

有多個測試案例，但不多於10組。
每個測試案例，只包含一個非負整型數字，代表詢問的年份y，0<=y<=4000000。

輸出

對於每個測試案例，如果是閏年輸出Y，如果不是閏年輸出N。

樣例輸入

0
2000
2002
2016

樣例輸出

N
N
Y
N

提示

來源

SL

這題是個啥玩意我還是沒明白

//C
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<map>
using namespace std;
const double eps = 1e-8;
int main(){
    int y;
    while(~scanf("%d",&y)){
        int i;
        double now = 0;
        for(i=1;i<=y;i++){
            now+=0.24219879;
            if(now>1.0){
                now -= 1.0;
            }
        }
        if(now+0.24219879>1||abs(now+0.24219879-1)<eps){
            printf("Y\n");
        }else{
            printf("N\n");
        }
        //cout<<now<<endl;
    }
    return 0;
}

1040: 世紀難題“晚飯吃啥”

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 23  解決: 12
[提交][狀態][討論版]

題目描述

   Zyyyyy和Song經常需要面對“晚飯吃啥”這個世紀難題，為了解決這個世紀難題，他們每次都需要一次比賽決定，比賽輸的一方來決定“晚飯吃啥”。他們比賽的題目是這樣的：

   現有“ABCDEFG”7種大寫字母，每個字母表示一個得分，“ABCDEFG”分別代表1、2、3、4、5、6、7分。給定一個只包含這7種字母的字串s，每人從s中選擇連續的n個字母，並將這n個字母代表的分數相加作為得分，得分多的一方獲勝。Song深知Zyyyyy智商爆表，所以她求助於你，希望你能寫個程式幫助她每次都能獲勝。

輸入

有T個測試案例，0<=T<=10。

每個測試案例有兩行輸入，第一行輸入兩個正整數m和n，分別表示字串的長度和選擇時字母連續的個數；第二行輸入一個長度為m的字串s。

0<m<=100，0<n<=m。輸入的字串保證只含有題目中描述的7種大寫字母。

輸出

對於每個測試案例，輸出最高的得分。

樣例輸入

3
7 3
ABCDEFG
7 3
DGAECBF
10 4
DFEGAGFDCB

樣例輸出

18
13
22

提示

來源

SL

//H
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;

int main(){
    int t,n,m,i;
    string in;
    //t = 'A';
    //cout<<t<<endl; 65
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&n);
        cin>>in;
        int now = 0,res = 0;
        for(i=0;i<n;i++){
            now += in[i]-64;
            //cout<<in[i] - 64;
            //cout<<now<<endl;
        }
        res = now;
        for(i=n;i<in.length();i++){
            now -= in[i-n]-64;
            //cout<<in[i-n];
            now += in[i]-64;
            //cout<<in[i]-64<<endl;
            res = max(res,now);
        }
        printf("%d\n",res);
    }
    return 0;
}

1041: A-B

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 18  解決: 8
[提交][狀態][討論版]

題目描述

    河海ACM隊一個盛產情侶的地方，比如大腿Song和巨腿Zyyyyy。Song和Zyyyyy的ACM之路都是從A+B這樣的水題開始的，然而他們早已厭倦了A+B這種無聊的題目，他們想來點改變，於是乎，想到了A-B問題。

    給出兩個八進位制的數字a,b，需要計算a-b的結果，也用8進製表示，如果結果是負數，你需要用負數形式輸出而不是給出補數。

輸入

第一行是一個整數T(1 <= T <= 1000), 代表資料組數。
每組陣列只有一行，為包括兩個八進位制的數a,b，其中0 <= a, b < 2^32。

輸出

輸出a-b的結果。

樣例輸入

1
76 7

樣例輸出

67

提示

來源

Song

//A
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<map>
using namespace std;
char res[100];
bool compare(string a,string b){
   if(a.length()>b.length()){
        return true;
   }else if(a.length()<b.length()){
       return false;
   }
   for(int i=0;i<=a.length();i++){
        if(a[i]>b[i]){
            return true;
        }else if(a[i]<b[i]){
            return false;
        }
   }
   return true;
}
int main(){
    int t,i;
    string a,b;
    scanf("%d",&t);
    while(t--){
        long long resa =0,resb = 0;
        //scanf("%lld%lld",&a,&b);
        cin>>a>>b;
        bool flag = false;
        if(!compare(a,b)){
            swap(a,b);
            flag = true;
        }
        int p = 0;
        int lena = a.length()-1;
        int lenb = b.length()-1;
        for(i=0;i<b.length();i++){
            if(a[lena-i]>=b[lenb-i]){
                res[++p] = a[lena-i] - b[lenb-i] ;
                //cout<<a[lena-i]<<" "<<b[lenb-i]<<endl;

            }else{
                res[++p] = 8 - (b[lenb-i]-'0') + a[lena-i]-'0';
                //cout<<b[lenb-i]-'0'<<" "<<a[lena-i]-'0'<<"   ";
                a[lena-i-1]--;
                //cout<<a[lena-i-1]<<endl;
            }
        }
        for(i=a.length()-b.length()-1;i>=0;i--){
            res[++p] = a[i]-'0';
            //cout<<"tes "<<a[i]<<endl;
        }
        //cout<<"p"<<p<<endl;
        while(res[p]==0&&p>=0){
                p--;
        }

        if(p==-1){
            printf("0\n");
            //cout<<"aaa\n";
            continue;
        }
        if(flag){
            printf("-");
        }
        for(i=p;i>=1;i--){
            printf("%d",res[i]);
        }
        printf("\n");
    }
    return 0;
}

1042: 特殊數

時間限制: 1 Sec  記憶體限制: 128 MB
提交: 15  解決: 6
[提交][狀態][討論版]

題目描述

    某天，Zyyyyy和Song想要一比智商高低，於是他們找到了河海ACM元老級先行者JetMuffin。JetMuffin給他們出了一道非常有意思的題目，是一道關於“特殊數”的題目。

一個“特殊數”就是出現的頻率最少的數字，給出一組整數，你需要輸出這組整數中的“特殊數”。注意如果有很多“特殊數”，你需要按順序輸出它們。

輸入

第一行是一個整數 T(1 <= T <= 4), 代表資料組數，
每一組資料第一行包括一個整數n (n <= 100000)，
第二行包括n個整數ai (0 <= a_i <= 10^9)。

輸出

每一組資料都按照升序輸出全部“特殊數”。

樣例輸入

2
7
1 1 2 2 3 3 3
5
5 3 2 4 1

樣例輸出

1 2
1 2 3 4 5

提示

注意最後一個數後面沒有空格。

來源

Song

//E
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<map>
using namespace std;
const int maxn = 100010;
int a[maxn];
int res[maxn];
map<int,int> judge;
int main(){
    int t,n,m,i,j,p;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        judge.clear();
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            judge[a[i]]++;
        }
        int minn = 1000000002;
        //cout<<"tes "<<minn<<endl;
        for(i=1;i<=n;i++){
            minn = min(judge[a[i]],minn);
        }
        j = 0;
        for(i=1;i<=n;i++){
            if(judge[a[i]]==minn){
                res[++j] = a[i];
            }
        }
        sort(res+1,res+1+j);
        for(i=1;i<=j;i++){
            printf("%d",res[i]);
            bool flag = false;
            while(i!=j&&res[i+1]==res[i]){
                i++;
                flag = true;
            }
            if(i!=j){
                printf(" ");
            }
        }
        printf("\n");
    }
    return 0;
}