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最新推荐文章于 2019-08-17 21:37:00 发布

原创最新推荐文章于 2019-08-17 21:37:00 发布 · 398 阅读

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本文介绍了一个字符算术谜题的解决方案，通过使用C++编程语言实现了一种算法来找出所有可能的字符到数字的映射方式，使得给定的字符算术表达式成立。该算法遍历所有数字组合，并检查字符串转换为数字后的算术关系是否满足等式。

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;
char s1[20], s2[20], s3[20];
int l1, l2, l3;
bool use[6];
int ok(int a[]) {
    int i, x, y, z, tmp;
    int ct[10];
    memset(ct, 0, sizeof(ct));
    for (i = 0; i < 5; i++) if (use[i]) ct[a[i]]++;
    for (i = 0; i <= 9; i++) if (ct[i] >= 2) return 0;
    x = y = z = 0;
    if ((a[s1[0] - 'A'] == 0 && l1 != 1) || (a[s2[0] - 'A'] == 0 && l2 != 1) || (a[s3[0] - 'A'] == 0 && l3 != 1)) return 0;
    tmp = 1;
    for (i = l1 - 1; i >= 0; i--) {
        x += a[s1[i] - 'A'] * tmp;
        tmp *= 10;
    }
    tmp = 1;
    for (i = l2 - 1; i >= 0; i--) {
        y += a[s2[i] - 'A'] * tmp;
        tmp *= 10;
    }
    tmp = 1;
    for (i = l3 - 1; i >= 0; i--) {
        z += a[s3[i] - 'A'] * tmp;
        tmp *= 10;
    }
    int sum = 0;
    if (x + y == z) sum++;
    if (x - y == z) sum++;
    if (x * y == z) sum++;
    if (y != 0 && x / y == z && x % y == 0) sum++;
    return sum;
}

int main() {
    int icase, i1, i2, i3, i4, i5;
    scanf("%d", &icase);
    while (icase--) {
        scanf("%s%s%s", s1, s2, s3);
        l1 = strlen(s1);
        l2 = strlen(s2);
        l3 = strlen(s3);
        memset(use, 0, sizeof(use));
        int i;
        for (i = 0; i < l1; i++) use[s1[i] - 'A'] = 1;
        for (i = 0; i < l2; i++) use[s2[i] - 'A'] = 1;
        for (i = 0; i < l3; i++) use[s3[i] - 'A'] = 1;
        int output = 0;
        for (i1 = 0; i1 <= 9; i1++) {
            if (use[0] == 0) i1 = 9;
            for (i2 = 0; i2 <= 9; i2++) {
                if (use[1] == 0) i2 = 9;
                for (i3 = 0; i3 <= 9; i3++) {
                    if (use[2] == 0) i3 = 9;
                    for (i4 = 0; i4 <= 9; i4++) {
                        if (use[3] == 0) i4 = 9;
                        for (i5 = 0; i5 <= 9; i5++) {
                            if (use[4] == 0) i5 = 9;
                            int a[6];
                            a[0] = i1, a[1] = i2, a[2] = i3, a[3] = i4, a[4] = i5;
                            output += ok(a);
                         }}}}}
        printf("%d\n", output);
    }
    return 0;
}