【递归练习】算24点

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代码

#include <cstdio>#include <cmath>#define EPS 1e-8using namespace std;double a[4];bool judge(double a[], int n){    if (n == 1){        if (fabs(a[0]-24) <= EPS) return 1;        else return 0;    }    double b[4];    for (int i = 0; i < n-1; ++i)        for (int j = i+1; j < n; ++j){            int rem = 0;            for (int k = 0; k < n; ++k)                if (k != i && k != j) b[rem++] = a[k];            b[rem] = a[i] + a[j];            if (judge(b, rem+1)) return 1;            b[rem] = a[i] - a[j];            if (judge(b, rem+1)) return 1;            b[rem] = a[j] - a[i];            if (judge(b, rem+1)) return 1;            b[rem] = a[i] * a[j];            if (judge(b, rem+1)) return 1;            if (fabs(a[j]) > EPS){                b[rem] = a[i] / a[j];                if (judge(b, rem+1)) return 1;            }            if (fabs(a[i]) > EPS){                b[rem] = a[j] / a[i];                if (judge(b, rem+1)) return 1;            }        }    return 0;}int main(){    while(~scanf("%lf%lf%lf%lf", &a[0], &a[1], &a[2], &a[3]) &&          (a[0] || a[1] || a[2] || a[3]))            judge(a, 4) ? printf("YES\n"): printf("NO\n");    return 0;}

说明

输入四个数，输出是否能算出24
允许出现分数/负数
比较浮点数是否相等不能用==

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