HUT-1674 火柴棒等式

该题算是一个暴力大表题了，给定了最多24根火柴棍，能够构成最大的数就是9992了，直接暴力。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;

int N, make[10] = {6,2,5,5,4,5,6,3,7,6}, A, B, C, cnt; 

int ans[30] = {
0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,8,9,6,9,29,39,38,65,88,128    
};

inline void fenjie(int num, int &sum)
{
    if (num == 0) {
        sum = 6;
        return;
    }
    sum = 0;
    while (num) {
        sum += make[num%10];
        num /= 10;
    }
}

void deal()
{
    int cnt_a, cnt_b, cnt_c;
    for (int i = 0; i < 9992; ++i) {
        fenjie(i, cnt_a);  // cnt_a的初始化在函数中执行 
        if (cnt_a >= N) {
            continue;
        }
        else { // 说明有足够多的火柴棍
            for (int j = 0; j <= 9992; ++j) {
                fenjie(j, cnt_b);
                if (cnt_b + cnt_a >= N) {
                    continue;
                }
                else {
                    fenjie(i+j, cnt_c);
                    if (cnt_c + cnt_a + cnt_b == N) {
                        ++cnt;
                    }
                }
            }
        }
    }
}

int main()
{
/*    for (int i = 0; i <= 24; ++i) {
        N = i;
        N -= 4; // 加号与等号各占去了4根火柴棍 
        cnt = 0;
        deal();
        printf("N = %d, ans = %d\n", N+4, cnt);
    } */ 
    while (scanf("%d", &N) == 1) { 
        printf("%d\n", ans[N]); 
    }
//    system("pause");
}