Codeforces 58E Expression (搜索)

题意：给你一个可能不正确的算式a + b = c, 你可以在a，b，c中随意添加数字。输出一个添加数字最少的新等式x + y  = z;

题目链接

思路：来源于这片博客：https://www.cnblogs.com/ljh2000-jump/p/5886279.html。

我们可以从个位开始搜索。如果现在c % 10 == （a % 10 + b % 10 + pre) % 10 (pre是之前的进位），那么直接把a, b, c的个位消去，加上进位继续深搜。如果不满足这个条件，我们可以考虑从a, b, c3个数中选一个数添加满足关系的一位，然后继续深搜。

代码:

#include <bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
int ans = INF;
LL ans_a, ans_b, ans_c;
LL p[20];
void dfs(LL a, LL b, LL c, LL nowa, LL nowb, LL nowc, LL pre, int now, int deep) {
	if(now >= ans) return;
	if(a == 0 && b == 0 && c == 0 && pre == 0) {
		ans = now;
		ans_a = nowa;
		ans_b = nowb;
		ans_c = nowc;
		return;
	}
	if(c == 0) {
		LL tmp = a + b + pre;
		int cnt = 0;
		while(tmp) {
			cnt++;
			tmp /= 10;
		}
		dfs(0, 0, 0, nowa + p[deep] * a, nowb + p[deep] * b, nowc + p[deep] * (a + b + pre), 0, now + cnt, deep + 1);
		return;
	}
	if((a + b + pre) % 10 == c % 10) {
		dfs(a / 10, b / 10, c / 10, nowa + (a % 10) * p[deep], nowb + (b % 10) * p[deep], nowc + (c % 10) * p[deep], (a %10 + b % 10 + pre) / 10, now, deep + 1);
		return; 
	}
	dfs(a * 10 + (c % 10 - b % 10 - pre + 10) % 10, b, c, nowa, nowb, nowc, pre, now + 1, deep);
	dfs(a, b * 10 + (c % 10 - a % 10 - pre + 10) % 10, c, nowa, nowb, nowc, pre, now + 1, deep);
	dfs(a, b, c * 10 + (a % 10 + b % 10 + pre) % 10, nowa, nowb, nowc, pre, now + 1, deep);
}
int main() {
	LL a, b, c;
	scanf("%lld+%lld=%lld", &a, &b, &c);
	p[0] = 1;
	for (int i = 1; i <= 18; i++)
		p[i] = p[i - 1] * 10;
	dfs(a, b, c, 0, 0, 0, 0, 0, 0);
	printf("%lld+%lld=%lld",ans_a, ans_b, ans_c);
}