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zoj3939 The Lucky Week 找规律

2017-04-03 14:41 323 查看
/*程序突然停止,数组太小 

(||)是怎么改变结果的?(110595 + 35502)%7  36524%7 = 5  

如果只有四年一闰, 4 (数的循环) * 7,

  如果只有四年一闰百年不闰,100(数的循环) * 7, 

  因为||400 % 4 == 0 改变了结果 (百年365 * 76 % 7 = 5   5 * 7 * 24 (百年)% 7 = 0

  ,四百年 365 *303 % 7 = 5 366 * 31 % 7 = 6 (5+5+5+6) % 7 = 0)*/

#include <iostream>

#include <cstdio>

using namespace std; 

void copysArray(int ans[13], int bs[13]);

void getPR(int p[13], int r[13]) {
copysArray(r, p);
r[2] = 29;

}

void copysArray(int ans[13], int b[13]) {
for (int i = 0; i < 13; i ++) {
ans[i] = b[i];
}

}

void getMounth(int mounth[13], int y) {
int p[13] = {
0,31,28,31,30,31,30,31,31,30,31,30,31
},r[13];
int originYear = 2016, year;
getPR(p, r);
year = y + originYear;

// copysArray(mounth, p);
if ((year % 4 ==0 && year % 100 != 0 ) || year % 400 == 0) {//r
copysArray(mounth, r);
} else {
copysArray(mounth, p);
}

}

void getData(int year[405][13][32],int ansYear[2060]){
int mounth[13];
bool inss = true;
int yt = 2000, days = 21, counts = 0;
year[2000 - yt][2][21] = 1;
for (int y = 0; y < 400; y ++) {
getMounth(mounth, y);
for (int m = 1; m <= 12; m ++) {
if (inss && y == 0) {
inss = false;
m = 2;
}
while (true){
if (days == 1 || days == 11 || days == 21) {
year[y][m][days] = ++counts;  //是记录
ansYear[counts] = y * 10000 + m * 100 + days;


days += 7;
//加7天  (* 一年365,7*52 = 364,没限制7年一循环)
if (days > mounth[m]) {// 超过一个月再下一个是几号 减这个月天数 
days -= mounth[m];
break;
//下一个月 
}
}
}
}  



int main(int argc, char *argv[])

{
int year[405][13][32] = {0};//m,d从一开始 
int ansYear[2060] = {0};
int y,m,d,n,t,ya,ym,yd,more4,sy;
getData(year, ansYear);
for (int i = 1; i < 2060; i++) {

// cout << i << "---" << ansYear[i] << endl;

while (cin >> y >> m >> d >> n) {
n--;
sy = ansYear[year[y%400][m][d]] / 10000;
t = year[y%400][m][d] + n;
more4 = t / 2058;
t = t % 2058;
ya = ansYear[t] / 10000 + y + more4 * 400 - sy;
ym = ansYear[t] % 10000;
ym = ym / 100;
yd = ansYear[t] % 100;
cout << ya << " " << ym << " " << yd << endl; 
}
return 0;
}
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