ZOJ 3939 The Lucky Week (暴力找规律)

题意：给定一个幸运日，求第 k 个幸运日是多少。

析：由于闰年，每400肯定会循环一次，所以我们就可以先找出每400年会有多少幸运日，是2058个，然后再暴力。

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 <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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;
}

int whatday(int d, int m, LL y){
  if(m == 1 || m == 2)  m += 12, --y;
  return (d + 2*m + 3*(m+1)/5 + y%7 + y/4%7 - y/100%7 + y/400%7) % 7 + 1;
}

int main(){
  int T;  cin >> T;
  while(T--){
    int y, m, d, kk;
    scanf("%d %d %d %d", &y, &m, &d, &kk);
    LL yy = kk / 2058 * 400LL + y;
    kk %= 2058;
    int ans = 0;
    for(LL i = yy; ; ++i){
      for(int j = (i == yy) ? m : 1; j < 13; ++j)
        for(int k = (i == yy && j == m) ? d : 1; k < 30; k += 10){
          if(whatday(k, j, i) == 1) ++ans;
          if(ans == kk){
            printf("%lld %d %d\n", i, j, k);
            goto A;
          }
        }
    }
    A:
    ;
  }
  return 0;
}