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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2734 Accepted Submission(s): 1010
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <stdlib.h> using namespace std; /*对于x=r0(mod m0) x=r1(mod m1) ... x=rn(mod mn) 输入数组m和数组r,返回[0,[m0,m1,...,mn]-1] 范围内满足以上等式的x0。 x的所有解为:x0+z*[m0,m1,...mn](z为整数) */ long long cal_axb(long long a,long long b,long long mod) { //防乘法溢出 long long sum=0; while(b) { if(b&1) sum=(sum+a)%mod; b>>=1; a=(a+a)%mod; } return sum; } //ax + by = gcd(a,b) //传入固定值a,b.放回 d=gcd(a,b), x , y void extendgcd(long long a,long long b,long long &d,long long &x,long long &y) { if(b==0){d=a;x=1;y=0;return;} extendgcd(b,a%b,d,y,x); y -= x*(a/b); } long long Multi_ModX(long long m[],long long r[],int n,long long &M) { long long m0,r0; m0=m[0]; r0=r[0]; for(int i=1;i<n;i++) { long long m1=m[i],r1=r[i]; long long k0,k1; long long tmpd; extendgcd(m0,m1,tmpd,k0,k1); if( (r1 - r0)%tmpd!=0 ) return -1; k0 *= (r1-r0)/tmpd; m1 *= m0/tmpd; r0 = ( cal_axb(k0,m0,m1)+r0)%m1; m0=m1; } M=m0; return (r0%m0+m0)%m0; } int main() { int T; cin>>T; int tt=1; while(T--) { int n; cin>>n; long long a[10],b[10]; for(int i=0;i<n;i++) cin>>a[i]; for(int i=0;i<n;i++) cin>>b[i]; long long M; long long ans=Multi_ModX(a,b,n,M); printf("Case %d: ",tt++); if(ans==0) ans+=M; cout<<ans<<endl; } return 0; }
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原文地址:http://www.cnblogs.com/chenhuan001/p/4993283.html