wenbao与中国剩余定理（孙子定理）

 

用来求解一般模线性方程，，

 

X %M1 == A1;

X %M2 == A2;

X %M3 == A3;

。。。。。

当M1, M2, M3，。。。互质时（关于不互质下面会提到），可以利用中国剩余定理求解。。

 

 

 

 

 

其中，而为模的逆元。

 

 

http://acm.hdu.edu.cn/showproblem.php?pid=1370

 

#include <stdio.h>
using namespace std;
int n, s[4], x, y, d, w, b[] = {23, 28, 33};
void exgcd(int a, int b, int& d, int& x, int& y){
    if(!b) d = a, x = 1, y = 0;
    else exgcd(b, a%b, d, y, x), y -= x*(a/b);
}
int solve(int *s){
    int M = 1, sum = 0, Mi;
    for(int i = 0; i < 3; i++){
        s[i]%=b[i];
        M*=b[i];
    }
    for(int i = 0; i < 3; i++){
        Mi = M/b[i];
        exgcd(Mi, b[i], d, x, y);
        sum = (sum + Mi*x*s[i]) % M;
    }
    sum = (M+sum%M)%M - w;
    if(sum <= 0) sum += 21252;
    return sum;
}
int main(){
    scanf("%d", &n);
    while(n--){
        int num = 1;
        while(scanf("%d %d %d %d", &s[0], &s[1], &s[2], &w)){
            if(s[0] == -1 && s[1] == -1 && s[2] == -1) break;
            printf("Case %d: the next triple peak occurs in %d days.\n", num++, solve(s));
        }
    }
    return 0;
}

 

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http://acm.fzu.edu.cn/problem.php?pid=1402

 

裸的剩余定理，，，，神奇的是fzu上面用VC可以过，但是用GNUC++却不可以，。。。。。。XXX

 

#include <iostream>
#include <stdio.h>
using namespace std;
#define ll long long
ll a[20], b[20], sum;
int t;
void ex(ll a, ll b, ll& x, ll& y){
    if(!b) x = 1, y = 0;
    else ex(b, a%b, y, x), y -= x*(a/b);
}
ll Ch(){
    ll xx = 0;
    for(int i = 0; i < t; ++i){
        ll sum2 = sum/a[i], x, y;
        ex(sum2, a[i], x, y);
        x = (x%a[i] + a[i])%a[i];
        xx = (xx + sum2*x*b[i]%sum)%sum;
    }
    return xx;
}
int main(){
    while(~scanf("%d", &t)){
        sum = 1;
        for(int i = 0; i < t; ++i){
            scanf("%lld%lld", &a[i], &b[i]);
            sum *= a[i];
        }
        printf("%lld\n", Ch());
    }
    return 0;
}

 

 

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http://poj.org/problem?id=1006

 

裸的剩余定理。。。。

 

#include <iostream>
using namespace std;
int p, e, i, d, sum = 21252, cnt = 0;
int a[3] = {924, 759, 644};
int b[3] = {23, 28, 33};
int c[3];
void xg(int a, int b, int &x, int &y){
    if(!b) x = 1, y = 0;
    else xg(b, a%b, y, x), y -= x*(a/b);
}
void Ch(){
    //cout<<23*28*33<<" "<<23*28<<" "<<23*33<<" "<<28*33<<endl;
    int num = 0, x, y;
    for(int i = 0; i < 3; ++i){
        xg(a[i], b[i], x, y);
        x = (x%b[i] + b[i])%b[i];
        num = (num + x*a[i]*c[i]%sum)%sum;
    }
    int xx = num - d;
    if(xx <= 0) xx += sum;
    printf("Case %d: the next triple peak occurs in %d days.\n", ++cnt, xx);
    //printf("%d\n", xx);
}
int main(){
    while(scanf("%d%d%d%d", &p, &e, &i, &d)){
        if(p == -1 && e == -1 && i == -1 && d == -1) break;
        c[0] = p % 23, c[1] = e % 28, c[2] = i % 33;
        Ch();
    }
    return 0;
}

 

 

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http://acm.hdu.edu.cn/showproblem.php?pid=5768

 

求X， Y之间减去（除pi余ai）后被7整除的数

剩余定理加容斥，。。。。涉及到枚举子集。。。

 

AC不易啊！！！！！

 

#include <iostream>
using namespace std;
#define ll long long
const int maxn = 20;
int n;
ll xxx, yyy, a[maxn], p[maxn];
int b[maxn];
void xg(ll a, ll b, ll &x, ll &y){
    if(!b) x = 1, y = 0;
    else xg(b, a%b, y, x), y -= x*(a/b);
}
ll Ch(int xx){
    int w;
    ll sum = 7LL, x, y, sum2, num = 0;
    for(int i = 0; i < xx; ++i){
        w = b[i];
        sum *= p[w];
    }
    for(int i = 0; i < xx; ++i){
        w = b[i];
        sum2 = sum/p[w];
        xg(sum2, p[w], x, y);
        x = (x%p[w]+p[w])%p[w];
        num = (num + x*sum2%sum*a[w]%sum)%sum;
    }
    return (yyy-num+sum)/sum - (xxx-num+sum)/sum;
}
ll solve(){
    ll sum = yyy/7 - xxx/7;
    ll len = 1LL << n;
    for(int i = 1; i < len; ++i){
        int num = 0;
        for(int j = 0; j < n; ++j){
            if(i&(1<<j)) b[num++] = j;
        }
        if(num&1) sum -= Ch(num);
        else sum += Ch(num);
    }
    return sum;
}
int main(){
    int t;
    scanf("%d", &t);
    for(int j = 1; j <= t; ++j){
        scanf("%d%lld%lld", &n, &xxx, &yyy);
        xxx--;
        for(int i = 0; i < n; ++i){
            scanf("%lld%lld", p+i, a+i);
        }
        printf("Case #%d: %lld\n", j, solve());
    }
    return 0;
}

 

 

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当M1, M2, M3.。。不互质时，该怎么求解呢？

通过合并方程可以求解，，

 

转载：

/**********************一般模线性方程组***********************/

同样是求这个东西。。
X mod m1=r1
X mod m2=r2
...
...
...
X mod mn=rn

首先，我们看两个式子的情况
X mod m1=r1……………………………………………………………(1)
X mod m2=r2……………………………………………………………(2)
则有 
X=m1*k1+r1………………………………………………………………(*)
X=m2*k2+r2
那么 m1*k1+r1=m2*k2+r2
整理，得
m1*k1-m2*k2=r2-r1
令(a,b,x,y,m)=(m1,m2,k1,k2,r2-r1)，原式变成
ax+by=m
熟悉吧？

此时，因为GCD(a,b)=1不一定成立，GCD(a,b) | m 也就不一定成立。所以应该先判 若 GCD(a,b) | m 不成立，则！！！方程无解！！！。
否则，继续往下。

解出(x,y)，将k1=x反代回（*），得到X。
于是X就是这两个方程的一个特解，通解就是 X'=X+k*LCM(m1,m2)
这个式子再一变形，得 X' mod LCM(m1,m2)=X
这个方程一出来，说明我们实现了(1)(2)两个方程的合并。
令 M=LCM(m1,m2)，R=r2-r1
就可将合并后的方程记为 X mod M = R。

然后，扩展到n个方程。
用合并后的方程再来和其他的方程按这样的方式进行合并，最后就能只剩下一个方程 X mod M=R，其中 M=LCM(m1,m2,...,mn)。
那么，X便是原模线性方程组的一个特解，通解为 X'=X+k*M。

如果，要得到X的最小正整数解，就还是原来那个方法：

X%=M;
if (X<0) X+=M;

 

 

例题

http://poj.org/problem?id=2891

 

 

#include <iostream>
using namespace std;
#define ll long long
const int maxn = 1e5+10;
ll p[maxn], a[maxn], t;
void ex(ll a, ll b, ll &d, ll &x, ll &y){
    if(!b) d = a, x = 1, y = 0;
    else ex(b, a%b, d, y, x), y -= x*(a/b);
}
ll solve(){
    ll b = p[0], c = a[0], d, x, y, M;
    for(int i = 1; i < t; ++i){
        ll xx = a[i] - c;
        ex(b, p[i], d, x, y);
        if(xx % d){
            return -1;
        }
        //cout<<x<<" "<<y<<"&&&"<<d<<endl;
        x = (xx/d*x)%(p[i]/d);
        //cout<<x<<"***"<<endl;
        c = b*x+c, b = b/d*p[i], c %= b;
        //cout<<c<<" "<<b<<endl;
    }
    return c <= 0 ? c+b : c;
}
int main(){
    while(~scanf("%lld", &t)){
        for(int i = 0; i < t; ++i){
            scanf("%lld%lld", p+i, a+i);
        }
        printf("%lld\n", solve());
    }
    return 0;
}

 

 

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http://acm.hdu.edu.cn/showproblem.php?pid=1573

中文题

 

 

 

 

#include <iostream>
using namespace std;
int n, m;
int p[11], a[11];
void ex(int a, int b, int &d, int &x, int &y){
    if(!b) d = a, x = 1, y = 0;
    else ex(b, a%b, d, y, x), y -= x*(a/b);
}
int solve(){
    int b = p[0], c = a[0], d, x, y;
    for(int i = 1; i < m; ++i){
        ex(b, p[i], d, x, y);
        int xx = a[i] - c;
        //cout<<i<<" "<<a[i]<<" "<<c<<" "<<xx<<" "<<d<<endl;
        if(xx%d){
            return 0;
        }
        x = xx/d*x;
        int w = p[i]/d;
        x = (x%w+w)%w;
        c +=x*b, b = b/d*p[i], c %= b;
    }
    c = (c > 0 ? c : c+b);
    if(c > n) return 0;
    else{
        return max((n-c)/b, 0)+1;
    }
}
int main(){
    int t;
    scanf("%d", &t);
    for(int i = 0; i < t; ++i){
        scanf("%d%d", &n, &m);
        for(int j = 0; j < m; ++j){
            scanf("%d", p+j);
        }
        for(int j = 0; j < m; ++j){
            scanf("%d", a+j);
        }
        printf("%d\n", solve());
    }
    return 0;
}

 

 

 

 

只有不断学习才能进步！