百度之星初赛A hdu6112

今夕何夕

题意：中文题

思路：每次加一年的天数，然后判模7是不是等于0，如果等于0就说明这年可行，要判时间是否合法，比如找到了2019-2-29就是不合法的，然后在2-29前的和2-29及其以后的有一点不一样，看代码把

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define ll long long
#define endl ("\n")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

int is[N];
void init(){
    for(int i=1; i<=10000; ++i){
        if(i%4==0 && i%100!=0) is[i]=366;
        else if(i%400==0) is[i]=366;
        else is[i]=365;
    }
}

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T;cin>>T;
    init();
    while(T--){
        int y,mm,dd,d=0,f=0,ff=0; char c;
        cin>>y>>c>>mm>>c>>dd;
        if(mm>2) f=1;
        if(mm==2 && dd==29) f=1, ff=1;
        for(int i=y; i<=10000; ++i){
            d+=is[i+f]; //cout<<is[i+f]<<"UUU\n";
            d%=7;
            if(ff){
                if(d==0 && is[i+1]==366){
                    cout<<i+1<<endl;
                    break;
                }
            }
            else if(d==0){
                cout<<i+1<<endl;
                break;
            }
        }
    }
    return 0;
}