中国剩余定理 poj1006

http://poj.org/problem?id=1006

裸的中国剩余定理 

x=a1mod(b1)

x=a2mod(b2)

x=a3mod(b3)

只需要求x1=a1*a2*i 使得a1*a2*i%a3==b3  等x2 x3 最后(x1+x2+x3)% lcm(x1,x2,x3)

 

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <set>#include <map>#include <vector>#include <queue>#include <fstream>#include <math.h>#include <iomanip>using namespace std;typedef long long ll;typedef pair<int,int> Pii;const int INF=0x7fffffff;int main(){    int x,y,z,d,cs=1;    while(cin>>x>>y>>z>>d){if(x==-1&&y==-1&&z==-1&&d==-1) break;x%=23;y%=28;z%=33;ll xx,yy,zz;for(int i=1;;i++){if(28*33*i%23==x) {xx=28*33*i;break;}}for(int i=1;;i++){if(33*23*i%28==y){yy=33*23*i;break;}}for(int i=1;;i++){if(28*23*i%33==z) {zz=28*23*i;break;}}ll t=xx+yy+zz-d+(23*28*33); t%=23*28*33;if(t==0) t+=23*28*33;printf("Case %d: the next triple peak occurs in %I64d days.\n",cs++,t);    }    return 0;}