史上最短12行题解

#include<bits/stdc++.h>
using namespace std;
int n,m[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int main()
{
	cin>>n;
	for(int i=1;i<=12;i++){
	if((12+n)%7==5)cout<<i<<endl;
	n = (m[i]+n)%7;
	}
	return 0;
}

可以直接模拟也可以用数学知识，判断x月的13日是不是星期五时，由于离上个月经过了（x-1）月的天数（1月另外算，+（13-1）天），就加（x-1）月的天数，再%7。

int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31},w;
	cin>>w;
	w+=12;
	for(int i=1;i<=12;i++){//12个月运行12次
		w%=7;
		if(w==5)cout<<i<<endl;//注意换行
		w+=a[i];
	}
	return 0;
}

#include <iostream>
using namespace std;
int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int main(){
    int n,w=0,num,j=12;
    cin>>n;
    num=n;
    while(j--){
        w++;
        for(int i=1;i<=a[w];i++,num++){
            if(num==8)
                num=1;
            if(i==13&&num==5){
                cout<<w<<endl;
            }
        }
    }
    return 0;
}

#include<bits/stdc++.h> using namespace std; int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int main(){ int n,w=0,num,j=12; cin>>n; num=n; while(j){ w++; for(int i=1;i<=a[w];i++,num++){ if(num8)num=1; if(i13&&num==5)cout<<w<<endl; } j--; } return 0; }

#include<bits/stdc++.h> using namespace std; int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int main(){ int n,w=0,num,j=12; cin>>n; num=n; while(j){ w++; for(int i=1;i<=a[w];i++,num++){ if(num8)num=1; if(i13&&num==5)cout<<w<<endl; } j--; } return 0; }

#include <math.h> #include <stdio.h> #include <iostream> #include <string.h> using namespace std; int main() { int l , r , d; cin >> l >> r >> d; int sum = 0; if(l < 2) l = 2; for(int k = 1 ; k <= r ; K++) { int flag = 0; int n = k; while(n) { if(n%10 == d) { flag = 1; break; } n/=10; } if(flag == 1) { n = k; for(int i = 2 ; i*i <= n ; i++) { if(n % i == 0) { flag = 0; break; } } if(flag == 1) sum++; } } cout << sum << endl; return 0; }

#include <math.h> #include <stdio.h> #include <iostream> #include <string.h> using namespace std; int main() { int l , r , d; cin >> l >> r >> d; int sum = 0; if(l < 2) l = 2; for(int k = 1 ; k <= r ; K++) { int flag = 0; int n = k; while(n) { if(n%10 == d) { flag = 1; break; } n/=10; } if(flag == 1) { n = k; for(int i = 2 ; i*i <= n ; i++) { if(n % i == 0) { flag = 0; break; } } if(flag == 1) sum++; } } cout << sum << endl; return 0; }