POJ 1006 Chinese Remainder Theory

https://www.youtube.com/watch?v=PM2D3xzqH_E
令 t mod 23 = a; t mod 28 = b; t mod 33 = c;
t= k * lcm(23, 28, 33) + x + y + z;
x= c1 * lcm (23, 28) 且 x mod 33 = c;  
y= c2 * lcm (23, 33) 且 y mod 28 = b;
z= c3 * lcm (28, 33) 且 z mod 23 = a;
因为 lcm(23, 28, 33) mod 23=0 mod 28=0 mod 33=0

所以，一步一步把mod加上去，具体见视频！

我的代码

#include<stdio.h>int x1,x2,x3,x4,a,b,c,d,e,f,g,x,y,z,nown,tot;int gcd(int a,int b){if (a%b==0)return b;int r=a % b;return gcd(b,r);}int lcm(int a, int b, int gcd){return gcd*a/gcd*b/gcd;}int main(){int a=23; int b=28; int c=33;int d=lcm(a,b,gcd(a,b)); int e=lcm(b,c,gcd(b,c));int f=lcm(a,c,gcd(a,c));int g=lcm(d, e, gcd(d,e));int casen=0;while (true){casen++;scanf("%d%d%d%d", &x1, &x2, &x3, &x4);if (x1==-1)break;int x= 1, y=1, z=1; //printf("d=%d e=%d f=%d\n", d,e,f);while (x*d%33!=x3%33)//有修改空间 //printf("x*d mod 33=%d\n",x*d%33);//getchar();//for(int i=0; i<=1e9; i++);x++;while (y*e%23!=x1%23)y++;while (z*f%28!=x2%28)z++;printf("x*d mod 33=%d x3 mod 33=%d\n", x*d % 33, x3 % 33);//printf("x=%d y=%d z=%d\n", x,y,z);tot=x*d+y*e+z*f;int nown=-tot/g; while (g*nown+tot<=x4 || g*nown+tot<x1 ||    g*nown+tot<x2 || g*nown+tot<x3)nown++;printf("Case %d: the next triple peak occurs in %d day", casen, g*nown+tot-x4);if (g*nown+tot-x4==1)printf(".\n");elseprintf("s.\n");}} 

别人的，比我牛逼。。

/*  中国剩余定理:出自《孙子算经》  */#include<stdio.h>#define MAX 21252int main(){int p, e, i, d, n, count = 0;while( scanf("%d%d%d%d", &p, &e, &i, &d) != EOF ){                count++;if(p == -1 && e == -1 && i == -1 && d == -1){break;                }n = ( 5544 * p + 14421 * e + 1288 * i - d ) % MAX;if( n <= 0 )   // 范围限制 {n += 21252;                }        printf("Case %d: the next triple peak occurs in %d days.\n", count, n );}return 0;}