中国剩余定理模板
互质 情况：
1 #include <cstdio>//互质 2 int exGcd(int a,int b,int &x,int &y) 3 { 4 if(b == 0) 5 { 6 x = 1,y = 0; 7 return a; 8 } 9 int d = exGcd(b,a%b,y,x); 10 y -= a/b*x; 11 return d; 12 } 13 int Chinese_Remainder(int mod[],int prime[],int len) 14 { 15 int i,d,x,y,m,n,ret; 16 ret = 0,n = 1; 17 for(i=0; i<len; i++) n *= prime[i]; 18 for(i=0; i<len; i++) 19 { 20 m = n/prime[i]; 21 d = exGcd(prime[i],m,x,y); 22 ret = (ret+y*m*mod[i])%n; 23 } 24 return (n+ret%n)%n; 25 } 26 int main() 27 { 28 int n,i; 29 int mod[15],prime[15]; 30 while(scanf("%d",&n)&&n) 31 { 32 for(i=0; i<n; i++) 33 scanf("%d%d",&prime[i],&mod[i]); 34 printf("%d\n",Chinese_Remainder(mod,prime,n)); 35 } 36 return 0; 37 }
不互质的情况：
1 //niiick 2 #include<iostream> 3 #include<vector> 4 #include<algorithm> 5 #include<queue> 6 #include<cstring> 7 #include<cstdio> 8 using namespace std; 9 typedef long long lt; 10 11 lt read() 12 { 13 lt f=1,x=0; 14 char ss=getchar(); 15 while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();} 16 while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();} 17 return f*x; 18 } 19 20 const int maxn=100010; 21 int n; 22 lt ai[maxn],bi[maxn]; 23 24 lt mul(lt a,lt b,lt mod) 25 { 26 lt res=0; 27 while(b>0) 28 { 29 if(b&1) res=(res+a)%mod; 30 a=(a+a)%mod; 31 b>>=1; 32 } 33 return res; 34 } 35 36 lt exgcd(lt a,lt b,lt &x,lt &y) 37 { 38 if(b==0){x=1;y=0;return a;} 39 lt gcd=exgcd(b,a%b,x,y); 40 lt tp=x; 41 x=y; y=tp-a/b*y; 42 return gcd; 43 } 44 45 lt excrt() 46 { 47 lt x,y,k; 48 lt M=bi[1],ans=ai[1];//第一个方程的解特判 49 for(int i=2;i<=n;i++) 50 { 51 lt a=M,b=bi[i],c=(ai[i]-ans%b+b)%b;//ax≡c(mod b) 52 lt gcd=exgcd(a,b,x,y),bg=b/gcd; 53 if(c%gcd!=0) return -1; //判断是否无解，然而这题其实不用 54 55 x=mul(x,c/gcd,bg); 56 ans+=x*M;//更新前k个方程组的答案 57 M*=bg;//M为前k个m的lcm 58 ans=(ans%M+M)%M; 59 } 60 return (ans%M+M)%M; 61 } 62 63 int main() 64 { 65 n=read(); 66 for(int i=1;i<=n;++i) 67 bi[i]=read(),ai[i]=read(); 68 printf("%lld",excrt()); 69 return 0; 70 }
