POJ 1006 中国剩余定理

【题意】：

给定p，e，i，d，求解

(x + d) % 23 = p 
(x + d) % 28 = e
(x + d) % 33 = i

x最小正整数值

【知识点】：

中国剩余定理

【题解】：

典型的 xmodmi = ai模型，其中mi间两两互素。但该题式子较少，也可以直接自己化简带入值。

【代码】：

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <ctime>
 5 #include <queue>
 6 #include <stack>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <bitset>
15 #include <climits>
16 #include <complex>
17 using namespace std;
18
19 #define wh while
20 #define inf (int)(~0u/2)
21 #define FOR(i, n) for(int i = 0; i < n; i++)
22 #define FOR1(i, n) for(int i = 1; i < n; i++)
23 #define FOR2(i, n) for(int i = 0; i <= n; i++)
24 #define REP(i,n) for(int i = 1; i <= n; i++)
25 #define FORI(it,n) for(typeof(n.begin()) it = n.begin(); it != n.end(); it++)
26 #define sf scanf
27 #define pf printf
28 #define frs first
29 #define sec second
30 #define psh push_back
31 #define mkp make_pair
32 #define PB(x) push_back(x)
33 #define MP(x, y) make_pair(x, y)
34 #define clr(abc,z) memset(abc,z,sizeof(abc))
35 #define lt(v) v << 1
36 #define rt(v) v << 1 | 1
37 #define mid ((l + r) >> 1)
38 #define lson l, mid, v << 1
39 #define rson mid + 1, r, v << 1 | 1
40 #define fre freopen("1.txt", "r", stdin)
41
42 typedef long long LL;
43 typedef long double LD;
44
45 LL exgcd(LL a, LL b, LL& x, LL& y){
46     if(b == 0){
47         x = 1; y = 0; return a;
48     }
49     else{
50         LL r = exgcd(b, a % b, y, x);
51         y -= x * (a / b);
52         return r;
53     }
54 }
55
56 int chinese(int a[], int m[], int n){
57     int M = 1, ret = 0;
58     for(int i = 0; i < n; i++) M *= m[i];
59     for(int i = 0; i < n; i++){
60         LL x, y;
61         int ta = M / m[i], tb = m[i];
62         exgcd(ta, tb, x, y);
63         ret = (ret + ta * x * a[i]) % M;
64     }
65     return (ret + M) % M;
66 }
67
68 int main(){
69     int p, e, i, d, it = 0;
70     while(sf("%d%d%d%d", &p, &e, &i, &d) != EOF){
71         if(p == -1 && e == -1 && i == -1 && d == -1) break;
72         int a[] = {p, e, i}, m[] = {23, 28, 33};
73         int ans = chinese(a, m, 3);
74         ans = (ans - d + 21252) % 21252;
75         if(ans == 0) ans = 21252;
76         pf("Case %d: the next triple peak occurs in %d days.\n", ++it, ans);
77     }
78 }

POJ 1006 中国剩余定理,布布扣,bubuko.com