uva 1588 换挡

http://acm.bnu.edu.cn/v3/contest_show.php?cid=5772#problem/H

给出两个每列高度只为1或2的长条，将它们放入一个高度为3的容器，求能够容纳它们的最短容器长度。

长度不超过100，枚举即可，分别固定条1和条2，另一个移动k单位看是否符合条件即可。

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <sstream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
const int inf = 1000000000;
const int maxn = 105;

char s1[maxn], s2[maxn];
int test(int k, char s1[], char s2[])
{
    for (int i = 0; s1[k+i] && s2[i]; i++)
        if (s1[k+i] - '0' + s2[i] - '0' > 3)
            return 0;
    return 1;
}
int fun(char s1[], char s2[])
{
    int k = 0;
    while (!test(k,s1,s2))
        k++;
    return max(strlen(s1), strlen(s2)+k);
}
int main()
{
    while (~scanf("%s%s", s1, s2))
        printf("%d\n", min(fun(s1,s2), fun(s2,s1)));
    return 0;
}