等式问题

等式问题

Time Limit:1s Memory Limit:1000k
Total Submit:2135 Accepted:1490
下载样例程序(PE)

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Problem

有一个未完成的等式1 2 3 4 5 6 7 8 9=N空格内（1前面没有空格）可以填入+,-,也可以不填。

编程找出输入某个整数 N 后使等式成立的所有方案的总数。保证有解。

Input

该题含有多组测试数据，每组数据为一行，仅一个整数N。

Output

输出方案总数。

Sample Input

108

Sample Output

15

C的求解和结果

#include <stdio.h>
#include <memory.h>

int main()
{
    char array[9];
    int N, count;
    int pos, result, i, temp, oper;
    while(scanf("%d", &N) > 0){
        memset(array, ' ', sizeof(array));
        count = 0;
        if(123456789 == N) count++;
        while(1){
            pos = 7;
            while(1){
                if(array[pos] == ' '){
                    array[pos] = '-';
                    break;
                }else if(array[pos] == '-'){
                    array[pos] = '+';
                    break;
                }else{
                    array[pos] = ' ';
                    pos--;
                    if(pos < 0) break;
                }
            }
            if(pos < 0) break;

            result = 0; temp = 1;   oper = 1;
            for(i = 0; i < 8; i++){
                if(array[i] == ' ') temp = temp * 10 + (i+2);
                else{
                    result += (oper * temp);
                    oper = (array[i] == '+') ? 1 : -1;
                    temp = i + 2;
                }
            }
            result += (oper * temp);
            if(result == N) count++;
        }
        printf("%d/n", count);
    }
    return 0;
}

Memory: 32K
Time: 7ms