中国剩余定理

先把板子留下，之后再来写证明…

Biorhythms HDU - 1370
除数互质

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;#define LL long longvoid gcd(LL a,LL b,LL &d,LL &x,LL &y){    if(b==0)    {        d=a;x=1;y=0;    }    else {        gcd(b,a%b,d,y,x);        y-=(a/b)*x;    }}int n=3;LL m[10],a[10];LL china(LL M){    LL d,y,z=0,x=0;    for(int i=0;i<n;i++)    {        LL w=M/m[i];        gcd(m[i],w,d,d,y);        x=(x+y*w*a[i])%M;    }    return (x+M)%M;}int main(){    int T;    scanf("%d",&T);    int d;    int M = 21252;    m[0]=23,m[1]=28,m[2]=33;    int case1=1;    while(scanf("%I64d %I64d %I64d %d",&a[0],&a[1],&a[2],&d)!=EOF)    {        if(a[0]==-1&&a[1]==-1&&a[2]==-1&&d==-1) break;        int c=china(M);        c-=d;        if(c<=0) c+=M;        printf("Case %d: the next triple peak occurs in %d days.\n",case1++,c);    }    return 0;}

除数不一定互质
X问题 HDU - 1573

#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#include <cstdio>#include <vector>using namespace std;const int maxn = 20;int a[maxn],n[maxn];int gcd(int aa,int bb){    if(bb==0) return aa;    else return gcd(bb,aa%bb);}void exgcd(int a1,int b1,int &g,int &x,int &y){    if(b1==0)    {        x=1;y=0;        g=a1;        return ;    }    else {        exgcd(b1,a1%b1,g,y,x);        y-=(a1/b1)*x;        return ;    }}bool merge1(int a1,int n1,int a2,int n2,int &aa,int &bb){    int d=gcd(n1,n2),k,x,y;    int c=a2-a1;    if(c%d) return false;    c=(c%n2+n2)%n2;    c/=d,n1/=d,n2/=d;    exgcd(n1,n2,k,x,y);    x=(x%n2+n2)%n2;    c*=x;c=(c%n2+n2)%n2;    c*=n1*d;c+=a1;    bb=n1*n2*d;    aa=(c%bb+bb)%bb;    return true;}int gg=0;int china(int len){    int a1=a[0],n1=n[0];    for(int i=1;i<len;i++)    {        int aa,nn;        int a2=a[i],n2=n[i];        if(!merge1(a1,n1,a2,n2,aa,nn)) return -1;        a1=aa,n1=nn;    }    gg=n1;    return (a1%n1+n1)%n1;}int main(){    int T;    scanf("%d",&T);    while(T--){        memset(a,0,sizeof(a));        memset(n,0,sizeof(n));        int nn,m;        scanf("%d %d",&nn,&m);        for(int i=0;i<m;i++)            scanf("%d",&n[i]);        for(int i=0;i<m;i++)            scanf("%d",&a[i]);        int k=china(m);       // cout<<"k="<<k<<endl;        if(k==-1||nn<k) printf("0\n");        else {            int num=(nn-k)/gg+1;            if(k==0)num--;            printf("%d\n",num);        }    }    return 0;}