poj 1006 Biorhythms(中國剩餘定理)
阿新 • • 發佈:2019-02-09
題意:
x === p ( mod 23 )
x === e ( mod 28 )
x === i ( mod 33 )
給p e i ,求x 。
解析:
23 28 33互素,所以用中國剩餘定理做。
用之前的解同餘方程組的方法也試了試,很噁心,樣例裡面有0 0 0 0 這樣的輸入,會爆除0錯。
程式碼:
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <cstring> #include <cmath> #include <stack> #include <vector> #include <queue> #include <map> #include <climits> #include <cassert> #define LL long long using namespace std; const int inf = 0x3f3f3f3f; const double eps = 1e-8; const double pi = 4 * atan(1.0); const double ee = exp(1.0); const int maxn = 30 + 10; void exgcd(int a, int b, int& d, int& x, int& y) { if (b == 0) { d = a; x = 1; y = 0; } else { exgcd(b, a % b, d, y, x); y -= a / b * x; } } int a[5], m[5], M; int CRT(int r) { M = 1; int res = 0; for (int i = 1; i <= r; i++) { M *= m[i]; } for (int i = 1; i <= r; i++) { int Mi = M / m[i]; int d, x, y; exgcd(Mi, m[i], d, x, y); res = (res + Mi * x * a[i]) % M; } if (res < 0) res += M; return res; } int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); #endif // LOCAL int ca = 1; int p, e, i, d; while (scanf("%d%d%d%d", &p, &e, &i, &d) == 4) { if (p + d + i + d == -4) break; a[1] = p, a[2] = e, a[3] = i; m[1] = 23, m[2] = 28, m[3] = 33; int ans = CRT(3); while (ans <= d) ans += M; printf("Case %d: the next triple peak occurs in %d days.\n", ca++, ans - d); } return 0; }