uva 185(回溯)

来源：互联网发布：scala java 混合使用编辑：程序博客网时间：2024/05/21 14:44

题意：给出了一些字母组成的等式，被看做是罗马数字的一个等式，先输出这个等式正确还是错误，然后把每个字母看做一个阿拉伯数字，然后判断这个等式，是有一种成立情况，还是多种，还是不可能成立。

题解：第一部分，只要让每一连串字母中它的每一个字母和下一个字母比较，如果当前一个比下一个大就是正数，否则就是负数，然后带入给出的表就能计算加判断了。第二部分有点麻烦，需要回溯+剪枝得到每一种可能排列情况。

#include <stdio.h>#include <string.h>#include <cmath>#include <string>#include <iostream>using namespace std;const int N = 10;const int M = 200;string str;int fir, sec, thr, map[M], flag, m[M], num, ln, rn, sn, flag1;char lef[N], rig[N], sum[N], app[N];int a, b, c;void init() {memset(m, 0, sizeof(m));flag = ln = rn = sn = num = flag1 = 0;memset(app, 0, sizeof(app));}int change(char *s, int len) {int temp = 0;for (int i = 0; i < len - 1; i++) {if (map[s[i]] >= map[s[i + 1]])temp += map[s[i]];elsetemp -= map[s[i]];}temp += map[s[len - 1]];return temp;}int change2(char *s, int len) {int temp = 0;int ans = 0;for (int i = len - 1; i >= 0; i--)ans += m[s[i]] * pow(10, temp++);//字符串转化整数return ans;}int judge() {a = change2(lef, ln);b = change2(rig, rn);c = change2(sum, sn);if (a + b == c)return 1;elsereturn 0;}void dfs(int cur) {if (flag == 2)return;if (cur == num) {if (judge())flag++;return;}for (int i = 0; i <= 9; i++) {if (!m[i]) {if (!i && (lef[0] == app[cur] || rig[0] == app[cur] || sum[0] == app[cur]))//排除前导零continue;if (flag1 != 1 && i > 1 && sum[0] == app[cur])//如果三个字符串长度相同，和大于1，否则都可以在和的首位是1的时候找到相应情况continue;m[app[cur]] = i;m[i] = 1;dfs(cur + 1);m[i] = 0;}}}int main() {int i, j;map['I'] = 1;map['X'] = 10;map['C'] = 100;map['M'] = 1000;map['V'] = 5;map['L'] = 50;map['D'] = 500;while (cin >> str && str[0] != '#') {init();int len = str.size();for (i = 0; str[i] != '+'; i++) {lef[ln++] = str[i];m[str[i]] = 1;}for (j = i + 1; str[j] != '='; j++) {rig[rn++] = str[j];m[str[j]] = 1;}for (int k = j + 1; k < len; k++) {sum[sn++] = str[k];m[str[k]] = 1;}for (i = 65; i <= 89; i++)if (m[i])app[num++] = i;fir = change(lef, ln);sec = change(rig, rn);thr = change(sum, sn);if (fir + sec == thr)printf("Correct ");elseprintf("Incorrect ");if (ln <= sn && rn <= sn) {if (ln == sn && rn == sn)flag1 = 1;dfs(0);}if (flag == 0)printf("impossible\n");else if (flag == 1)printf("valid\n");elseprintf("ambiguous\n");}return 0;}

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