PAT乙級1011-1020
阿新 • • 發佈：2018-12-24
1011A+B和C，
注意要使用longlong型別，因為兩個32位相加有可能超出範圍導致符號反轉
#include<iostream> #include<vector> using namespace std; int main(int argc, char*argv[]) { int sum; cin >> sum; vector<vector<long long> >arrays; for (int i = 0; i < sum; i++) { vector<long long>a(3); cin >> a[0] >> a[1] >> a[2]; arrays.push_back(a); } for (int i = 0; i < arrays.size(); i++) { cout << "Case #" << i + 1 << ": "; if (arrays[i][0] + arrays[i][1] > arrays[i][2]) cout << "true" << endl; else cout << "false" << endl; } return 0; }
1012，數字分類,
注意演算法庫中函式的使用，求和，求最大值，求積，注意格式化輸出一個小數
#include<iostream> #include<vector> #include<numeric> #include<functional> #include<iomanip> #include<algorithm> #include"stdlib.h" using namespace std; int main(int argc, char*argv[]) { int sum; cin >> sum; vector<int>nums(sum); for (int i = 0; i < sum; i++) cin >> nums[i]; vector<vector<int> >clfs(5); for (auto i : nums) { if (i % 5 == 0) { if (i % 2 == 0) clfs[0].push_back(i); } else clfs[i % 5].push_back(i); } for (int j = 0; j < clfs.size(); j++) { int sig = 1; int sum = 0; double avr; if (clfs[j].size() != 0) { switch (j) { case 0: cout << accumulate(clfs[j].begin(), clfs[j].end(),0); break; case 1: for (auto i : clfs[j]) { sum += sig*i; sig *= -1; } cout << " " << sum; break; case 2: cout << " " << clfs[j].size(); break; case 3: avr = accumulate(clfs[j].begin(), clfs[j].end(), 0.0) / clfs[j].size(); cout << " " << setiosflags(ios::fixed) << setprecision(1) << avr; break; case 4: cout << " " << *(max_element(clfs[j].begin(), clfs[j].end())); break; default: ; } } else { if (j == 0) cout <<'N'; else cout <<" N"; } } return 0; }
1013，數素數
注意按規則輸出數字
#include<iostream> #include<vector> #include<math.h> using namespace std; bool is_prim(int a); int main(int argc, char*argv[]) { int begin = 0; int end = 0; cin >> begin >> end; int i = 0; int num = 2; vector<int>prims; while (i <= end) { if (is_prim(num)) { prims.push_back(num); i++; } num++; } cout << prims[begin - 1]; for (int i = begin; i < end; i++) { if ((i - begin) % 10 == 9) cout << "\n"; else cout << " "; cout << prims[i]; } return 0; } bool is_prim(int a) { for (int i = 2; i <= sqrt(a); i++) if (a%i == 0) return false; return true; }
1014，福爾摩斯的約會,(非原創)
注意格式化輸出字串。注意char字元與int型別的ascii碼的對換
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
char Day[7][4] = {"MON","TUE","WED","THU","FRI","SAT","SUN"};
int flag(0) , H(0) , M(0) ,cnt(0);
char L,DAY='A';
string A,B,C,D;
cin >> A >> B >> C >> D;
int MinF = A.length() < B.length() ? A.length() : B.length();
int MinT = C.length() < D.length() ? C.length() : D.length();
for(int i=0;i<MinF;i++){
if(A[i] == B[i] && flag == 0 && A[i] >= 'A' && A[i]<= 'G'){
L=A[i];
flag = 1;
continue;
}
if(A[i]==B[i] && A[i] - '0' >= 0 && A[i]-'0'<=9 && flag==1){
H = A[i]-'0';
break;
}else if(A[i]==B[i]&&A[i]>='A'&&A[i]<='N'&&flag==1){
H = A[i] - 'A' +10; //×¢Òâ
break;
}
}
cnt = L- 'A';
for(int i=0;i<MinT;i++){
if(C[i]==D[i]&& ( (C[i]>='A'&&C[i]<='Z') || (C[i]>='a' && C[i]<='z') ) ){ //×¢Òâ
M=i;
break;
}
}
cout << Day[cnt];
printf(" %02d:%02d",H,M);
return 0;
} 注意使用algrothm庫中的sort函式來進行自定義排序
注意使用scanf(在ide中是scanf_s)來解決執行超時問題，不要用iostream
#include<cstdio>
#include<vector>
#include<algorithm>
struct stu {
int num;
int de;
int cai;
int lei;
};
bool compare(stu a,stu b);
using namespace std;
int main(int argc, char*argv[])
{
int N, L, H;
scanf_s("%d %d %d", &N, &L, &H);
vector<stu>stus(N);
int M=0;
for (int i = 0; i < N; i++)
{
//cin >> stus[i].num >> stus[i].de >> stus[i].cai;
scanf_s("%d %d %d", &stus[i].num, &stus[i].de, &stus[i].cai);
if (stus[i].de >= H&&stus[i].cai >= H)
stus[i].lei = 0;
else if (stus[i].de >= H&&stus[i].cai < H&&stus[i].cai >= L)
stus[i].lei = 1;
else if (stus[i].de < H&&stus[i].de >= L&&stus[i].cai < H&&stus[i].cai >= L&&stus[i].de>=stus[i].cai)
stus[i].lei = 2;
else if (stus[i].de < H&&stus[i].de >= L&&stus[i].cai >= L&&stus[i].de < stus[i].cai)
stus[i].lei = 3;
else
{
stus[i].lei = 4;
M++;
}
}
sort(&stus[0], &stus[0] + N, compare);
printf("%d\n", N - M);
for (int i = 0; i < N - M; i++)
printf("%d %d %d\n", stus[i].num, stus[i].de, stus[i].cai);
//cout << stus[i].num << " " << stus[i].de << " " << stus[i].cai << endl;
return 0;
}
bool compare(stu a, stu b)
{
if (a.lei < b.lei)
return true;
else if (a.lei>b.lei)
return false;
else {
if (a.de + a.cai > b.de + b.cai)
return true;
else if (a.de + a.cai < b.de + b.cai)
return false;
else {
if (a.de > b.de)
return true;
else if (a.de < b.de)
return false;
else
return a.num < b.num;
}
}
}採用求餘遍歷來找數字
#include<iostream>
using namespace std;
int main(int argc, char*argv[])
{
long long A, B;
int Da, Db;
cin >> A >> Da >> B >> Db;
int Pa = 0;
int Pb = 0;
int ina = 1;
int inb = 1;
while (A)
{
if (A % 10 == Da)
{
Pa += Da*ina;
ina *= 10;
}
A /= 10;
}
while (B)
{
if (B % 10 == Db)
{
Pb += Db*inb;
inb *= 10;
}
B /= 10;
}
cout << Pa + Pb;
return 0;
}1017A除以B（非原創）
注意最好不要分段計算，而是逐位的計算並輸出商數和餘數
#include <iostream>
#include <string>
using namespace std;
int main(){
string s;
cin >> s;
int a;
cin >> a;
int len = s.length();
int t = (s[0]-'0')/a;
if((t!=0&&len>1) || len==1){
cout << t;
}
int temp = (s[0]-'0') % a;
for(int i=1;i<len;i++){
t = (temp*10 + s[i]-'0')/a;
cout << t;
temp = (temp*10 + s[i]-'0')%a;
}
cout << " " << temp;
return 0;
}我的辦法既愚蠢又笨拙，有更好的方法嗎，
#include<iostream>
using namespace std;
int main(int argc, char*argv[])
{
int win(0), loss(0), ping(0);
int jia[3] = { 0,0,0 };
int yi[3] = { 0,0,0 };
int num;
cin >> num;
char a, b;
for (int i = 0; i < num; i++)
{
cin >> a >> b;
switch (a - b)
{
case 0:
ping++; break;
case 1:
yi[0]++; loss++; break;
case -1:
jia[0]++; win++; break;
case 8:
jia[2]++; win++; break;
case -8:
yi[2]++; loss++; break;
case 7:
yi[1]++; loss++; break;
case -7:
jia[1]++; win++; break;
default:
;
}
}
cout << win << " " << ping << " " << loss << endl;
cout << loss << " " << ping << " " << win << endl;
int a1(0), a2(0);
int b1(0), b2(0);
for (int i = 0; i < 3; i++)
{
if (jia[i] > a1)
{
a2 = i; a1 = jia[i];
}
if (yi[i] > b1)
{
b2 = i; b1 = yi[i];
}
}
switch (a2)
{
case 0:
cout << "B "; break;
case 1:
cout << "C "; break;
case 2:
cout << "J "; break;
default:;
}
switch (b2)
{
case 0:
cout <<"B"; break;
case 1:
cout <<"C"; break;
case 2:
cout <<"J"; break;
default:;
}
}1019數字黑洞（非原創）
注意stoi函式和string的insert函式的使用
#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(char a,char b){return a>b;}
int main(){
string s;
cin >> s;
s.insert(0,4-s.length(),'0');
do{
string a=s,b=s;
sort(a.begin(),a.end(),cmp);
sort(b.begin(),b.end());
int result = stoi(a)-stoi(b);
s = to_string(result);
s.insert(0,4-s.length(),'0');
cout << a << " - " << b << " = " << s << endl;
}while(s!="6174" && s!="0000");
return 0;
}注意輸出格式即可，還有就是最好避免強制型別轉換，而是一開始就用float
#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
using namespace std;
struct goods {
float s;
float price;
};
bool compare(goods a, goods b)
{
return a.price / a.s > b.price / b.s;
}
int main(int argc, char*argv[])
{
int sum, max;
cin >> sum >> max;
vector<goods>stock(sum);
for (int i = 0; i < sum; i++)
cin >> stock[i].s;
for (int i = 0; i < sum; i++)
cin >> stock[i].price;
sort(stock.begin(), stock.end(), compare);
float maxprofit = 0;
for (int i = 0; i < sum; i++)
{
if (stock[i].s <= max)
maxprofit += stock[i].price;
else
{
maxprofit += stock[i].price / stock[i].s*max;
break;
}
max -= stock[i].s;
}
cout <<setiosflags(ios::fixed)<<setprecision(2)<< maxprofit;
return 0;
}