POJ 1061 青蛙的约会（扩展欧几里德）

标签：扩展   gcd   扩展欧几里德   

题目链接：http://poj.org/problem?id=1061

Description

Input

Output

Sample Input

1 2 3 4 5

Sample Output

4

Source

PS:

根据题意得：x + m*t - (y + n*t) = k*l;  ---->>  t*(m-n) - k*l = y-x  --->>  k*l + t*(n-m) = x-y;

最终化简得：k * l + t*(n - m) = x - y;

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
typedef __int64 LL;

LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    else
    {
        LL r = exgcd(b,a%b,x,y);
        LL t = x;
        x = y;
        y = t-a/b*y;
        return r;
    }
}

LL cal(LL a, LL b, LL c)
{
    LL x, y;
    LL tt = exgcd(a, b, x, y);
    if(c%tt)//无整数解
    {
        return -1;
    }
    x*=c/tt;
    b/=tt;
    LL ans = (x%b+b)%b;//最小整数解
    return ans;
}

int main()
{
    LL a, b, n, m, l, x, y;
    while(~scanf("%I64d%I64d%I64d%I64d%I64d",&x,&y,&m,&n,&l))
    {
        //t(n-m) + k*l = x - y;
        LL ans = cal(n-m, l, x-y);
        if(ans == -1)
        {
            printf("Impossible\n");
            continue;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

POJ 1061 青蛙的约会（扩展欧几里德）

标签：扩展   gcd   扩展欧几里德   

原文地址：http://blog.csdn.net/u012860063/article/details/39778135