标签:ota 同余 auth nes other 整数 clu problems har
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3174 Accepted Submission(s): 1371
#include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define N 20 #define ll long long using namespace std; ll n,m[N],a[N],m1,e; ll read() { ll x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();} return x*f; } ll exgcd(ll a,ll b,ll &x,ll &y) { if(b==0) { x=1,y=0; return a; } ll r=exgcd(b,a%b,x,y),tmp; tmp=x,x=y,y=tmp-a/b*y; return r; } ll crt() { ll a1=a[1],a2,m2,d,c;m1=m[1]; for(ll i=2;i<=n;++i) { a2=a[i],m2=m[i]; c=a2-a1;ll x=0,y=0; d=exgcd(m1,m2,x,y); if(c%d) return -1; x=x*c/d; int mod=m2/d; x=(mod+x%mod)%mod; a1+=m1*x;m1*=mod; } return a1; } int main() { while(1) { n=read(),e=read(); if(n==0&&e==0) break; for(int i=1;i<=n;i++) m[i]=read(),a[i]=m[i]-e; printf("%lld\n",crt()); } return 0; }
HDU——1788 Chinese remainder theorem again
标签:ota 同余 auth nes other 整数 clu problems har
原文地址:http://www.cnblogs.com/z360/p/7351619.html