CCF日期计算，解题思路，C++,Java

解题思路：

首先判断是闰年还是平年，若是瑞年则28+1，平年28+0

bool flag = n%4 ==0&& n%100 != 0 || n%400 == 0 ;if(flag){t=1;}else{t=0;}

#include <iostream>using namespace std;int main(){int n,d,t;cin>>n>>d;if(d <=31){cout<<1<<endl;cout<<d;return 0;}bool flag = n%4 ==0&& n%100 != 0 || n%400 == 0 ;if(flag){t=1;}else{t=0;}if( d <=59+t){// 31+28 + t 如果t=1 则 d< 60 即2月有29天，若是0，则2月有28天 cout<<2<<endl;cout<< d-31;}else if( d <= 90+t){ // 31 + 2月（28 + t）+ 31 = 90+t  如果t=1 则 d< 60 即2月有29天，若是0，则2月有28天 ,下面的情况类似 cout<<3 <<endl;cout<< d-(59+t);}else if( d <= 120+t){cout<<4<<endl;cout<< d-(90+t);}else if( d <= 151+t){cout<<5 <<endl;cout<< d-120-t;}else if( d <= 181+t){cout<<6 <<endl;cout<< d-151-t;}else if( d <= 212+t){cout<<7<<endl;cout<< d-181-t;}else if( d <= 243+t){cout<<8 <<endl;cout<< d-212-t;}else if( d <= 273+t){cout<<9 <<endl;cout<< d-243-t;}else if( d <= 304+t){cout<<10 <<endl;cout<< d-273-t;}else if( d <= 334+t){cout<<11 <<endl;cout<< d-304-t;}else if( d <= 365+t){cout<<12 <<endl;cout<< d-334-t;}return 0;}