A research laboratory of a world-leading automobile company has received an order to create a special transmission mechanism, which allows for incredibly efficient kickdown — an operation of switching to lower gear. After several months of research engineers found that the most efficient solution requires special gears with teeth and cavities placed non-uniformly. They calculated the optimal flanks of the gears. Now they want to perform some experiments to prove their findings. The first phase of the experiment is done with planar toothed sections, not round-shaped gears. A section of length n consists of n units. The unit is either a cavity of height h or a tooth of height 2h. Two sections are required for the experiment: one to emulate master gear (with teeth at the bottom) and one for the driven gear (with teeth at the top). There is a long stripe of width 3h in the laboratory and its length is enough for cutting two engaged sections together. The sections are irregular but they may still be put together if shifted along each other. The stripe is made of an expensive alloy, so the engineers want to use as little of it as possible. You need to find the minimal length of the stripe which is enough for cutting both sections simultaneously. Input The input file contains several test cases, each of them as described below. There are two lines in the input, each contains a string to describe a section. The first line describes master section (teeth at the bottom) and the second line describes driven section (teeth at the top). Each character in a string represents one section unit — 1 for a cavity and 2 for a tooth. The sections can not be flipped or rotated. Each string is non-empty and its length does not exceed 100. Output For each test case, write to the output a line containing a single integer number — the minimal length of the stripe required to cut off given sections. Sample Input 2112112112 2212112 12121212 21212121 2211221122 21212 Sample Output 10 8 15
字符串问题，匹配的时候，只要不同时为2 就可以，并且要注意，可以移动上边的，也可以移动下边的，但是不可以翻转~
#include <iostream> #include<cstdio> #include<cstring> using namespace std; #define MAXN 12345672 char p[MAXN]; char t[MAXN]; int getlens(char t[], char p[]) { int flag; int len1 = strlen(t); int len2 = strlen(p); for(int i=0;i<len1;i++)//移动 { flag = 1; for(int j=i;j<len1&&j<i+len2;j++)//匹配 { if(t[j]=='2'&&p[j-i]=='2') flag = 0; } if(flag) return max(len1, i+len2);//如果第一个字符串的长度比匹配之后的i+len2还要长，就要返回len1，所以是返回最大的 } return len1 + len2; } int main() { while(~scanf("%s", p)) { scanf("%s", t); int len = min(getlens(t, p), getlens(p, t));//找到长度最小的 printf("%d\n", len); } return 0; }