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题意：给定两个日期，两种不同算闰年的方法，导致日期不同，给定那个慢的，求你求了那个快的。
析：因为算闰年的方法不同，所以我们就要先从1582算到当前时间，算出差了多少天，再加上就好。注意跨月，跨年的情况。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
bool is_g(int y){
if(y % 400 == 0) return true;
if(y % 100 == 0) return false;
if(y % 4 == 0) return true;
return false;
}
bool is_j(int y){
return y % 4 == 0;
}
bool f(int a, int b, int c)
{
int k = 28;
if(is_g(a)) k = 29;
if(b == 2 && c <= k) return true;
else if((b == 1 || b == 3 || b == 5 || b == 7 || b == 8 || b == 10 || b == 12) && (c <= 31)) return true;
else if((b == 4 || b == 6 || b == 9 || b == 11) && (c <= 30)) return true;
return false;
}
int cntj[10005];
int cntg[10005];
void solve(int y, int m, int d){
int cnt = cntj[y-1] - cntg[y-1];
if(m > 2 || (m == 2 && d > 28)) cnt += is_j(y) - is_g(y);
cnt += 11;
while(cnt--)
{
d++;
if(f(y, m, d)) continue;
d = 1; m++;
if(f(y, m, d)) continue;
y++; m = 1;
}
printf("%d-%02d-%02d\n", y, m, d);
}
int main(){
int y, m, d;
for(int i = 1584; i <= 10000; i += 4){
++cntj[i];
cntg[i] += is_g(i);
}
for(int i = 1584; i <= 10000; ++i)
cntj[i] += cntj[i-1], cntg[i] += cntg[i-1];
while(~scanf("%d-%d-%d", &y, &m, &d)){
solve(y, m, d);
}
}
LA UVaLiva 6627 First Date (转换时间)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5791596.html
