中国剩余定理例题

最新推荐文章于 2023-05-14 13:35:06 发布

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最新推荐文章于 2023-05-14 13:35:06 发布

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CC 4.0 BY-SA版权

文章标签：密码学 c/c++

原文链接：http://www.cnblogs.com/wzq--boke/p/9772823.html

本文精选了多个算法竞赛中的经典题目代码实现，包括荒岛野人、五指山、Biorhythms、CLooooops等，涵盖了暴力枚举、中国剩余定理、扩展欧几里得算法等核心算法。

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暴力枚举山洞数，再n^2两两判断是否成立

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
const int maxn=1000000;
using namespace std;
typedef long long ll;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=(x<<3)+(x<<1)+ch-'0';
    return x*f;
}
int n,c[20],p[20],l[20],st;
 
 
ll exgcd(int a,int b,int &x,int &y){
    if (b==0){x=1,y=0;return a;}
    int z=exgcd(b,a%b,x,y),t=x;
    x=y,y=t-a/b*y;return z;
}
 
bool check(int k){
    for(int i=1;i<n;i++){
        for(int j=i+1;j<=n;j++){
            int W=c[j]-c[i],Z=p[i]-p[j];
            int d,x,y;
            d=exgcd(Z,k,x,y);
            if(W%d)continue;
            x=(x*W/d)%(k/d);
            if(x<0)x+=abs(k/d);
            if(x<=l[i] && x<=l[j])return 0;
        }
    }
    return 1;
}
 
int main(){
    n=read();
    for(int i=1;i<=n;i++)scanf("%d%d%d",&c[i],&p[i],&l[i]),st=max(st,c[i]);
    for(int i=st;i<=1000000;i++){
        if(check(i)){printf("%d",i);break;}
    }
    return 0;
}

裸地crt

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define ll long long
using namespace std;
const int maxn=100000+10101;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=(x<<3)+(x<<1)+ch-'0';
    return x*f;
}

void exgcd(int a,int b,int &d,ll &x,ll &y){
    if(b==0){
        d=a;
        x=1;y=0;
        return ;
    }
    exgcd(b,a%b,d,x,y);
    ll t=x;x=y;
    y=t-a/b*y;
    return ;
}

int t;



int main(){
    t=read();
    for(int i=1;i<=t;i++){
        int n=read(),d=read(),x=read(),y=read();
        y=y-x;
        int gcd;
        ll x1,y1;
        exgcd(d,n,gcd,x1,y1);
        if(y%gcd){printf("Impossible\n");continue;}
        x1=((x1*(y/gcd))%(n/gcd)+n/gcd)%(n/gcd);
        printf("%lld\n",x1);
    }
    return 0;
}

因为23,28,33互质，所以用crt就行了

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define ll long long
using namespace std;
const int maxn=1000000+10101;
inline int read(){
    int x=0,f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=(x<<3)+(x<<1)+ch-'0';
    return x*f;
}

int d;
int m[3]={23,28,33},r[3],M=21252;

void exgcd(int a,int b,int &x,int &y){
    if(!b){
        x=1;y=0;
        return ;
    }
    exgcd(b,a%b,x,y);
    int t=x;x=y;
    y=t-a/b*y;
    return ;
}
int main(){
    int cnt=0;
    while(1){
        r[0]=read();r[1]=read();r[2]=read();d=read();
        if(d==r[0] && r[0]==r[1] && r[1]==r[2] && d==-1)break;
        cnt++;
        int ans=0;
        for(int i=0;i<3;i++){
            int ww=M/m[i];
            int x,y;
            exgcd(ww,m[i],x,y);
            ans=((ans+r[i]*ww*x)%M+M)%M;
        }
        if(ans<=d)ans+=M;
        printf("Case %d: the next triple peak occurs in %d days.\n",cnt,ans-d);
    }
    return 0;
}

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define ll long long
using namespace std;
const int maxn=1000000+10101;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=(x<<3)+(x<<1)+ch-'0';
    return x*f;
}

ll A,B,C,K;
ll d,x,y;
void exgcd(ll a,ll b,ll &d,ll &x,ll &y){
    if(!b){
        d=a;x=1;y=0;
        return ;
    }
    exgcd(b,a%b,d,x,y);
    ll t=x;x=y;
    y=t-a/b*y;
    return ;
}
int main(){
    while(1){
        A=read();B=read();C=read();K=read();
        if(A==0 && B==0 && C==0 && K==0)break;
        ll sum=(1ll<<K);
        B=((B-A)%sum+sum)%sum;
        exgcd(C,sum,d,x,y);
        if(B%d==0){
            sum=sum/d;
            x=((x*B/d)%sum+sum)%sum;
            printf("%lld\n",x);        
        }
        else printf("FOREVER\n");
    }
    return 0;
}