Justice String
Time Limit: 2000ms
Memory Limit: 65536KB
64-bit integer IO format: %lld Java class name: MainGiven two strings A and B, your task is to find a substring of A called justice string, which has the same length as B, and only has at most two characters different from B.
Input
The first line of the input contains a single integer T, which is the number of test cases.
For each test case, the first line is string A, and the second is string B.
Both string A and B contain lowercase English letters from a to z only. And the length of these two strings is between 1 and 100000, inclusive.
Output
For each case, first output the case number as "Case #x: ", and x is the case number. Then output a number indicating the start position of substring C in A, position is counted from 0. If there is no such substring C, output -1.
And if there are multiple solutions, output the smallest one.
Sample Input
3 aaabcd abee aaaaaa aaaaa aaaaaa aabbb
Sample Output
Case #1: 2 Case #2: 0 Case #3: -1
Source
解题：Hash+二分
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 100100; 18 char sa[maxn],sb[maxn]; 19 unsigned LL Ha[maxn],Hb[maxn],p = 131; 20 int len1,len2; 21 unsigned LL pp[maxn]; 22 int calc(int a,int b){ 23 int high = min(len1 - a,len2 - b),low = 0,mid,ans = 0; 24 while(low <= high){ 25 mid = (low + high)>>1; 26 unsigned LL v1 = Ha[a] - Ha[a+mid]*pp[mid]; 27 unsigned LL v2 = Hb[b] - Hb[b+mid]*pp[mid]; 28 if(v1 == v2){ 29 ans = mid; 30 low = mid + 1; 31 }else high = mid - 1; 32 } 33 return ans; 34 } 35 int main() { 36 int T,cs = 1; 37 pp[0] = 1; 38 for(int i = 1; i < maxn; ++i) pp[i] = pp[i-1]*p; 39 scanf("%d",&T); 40 while(T--){ 41 scanf("%s %s",sa,sb); 42 len1 = strlen(sa); 43 len2 = strlen(sb); 44 Ha[len1] = 0; 45 Hb[len2] = 0; 46 for(int i = len1-1; i >= 0; --i) Ha[i] = Ha[i+1]*p + sa[i]; 47 for(int i = len2-1; i >= 0; --i) Hb[i] = Hb[i+1]*p + sb[i]; 48 int ans = -1; 49 for(int i = 0; i <= len1 - len2; ++i){ 50 int sum = 0,a = i,b = 0,tmp = 0; 51 tmp = calc(a,b); 52 if(tmp >= len2-2){ 53 ans = i; 54 break; 55 } 56 sum += tmp; 57 a += tmp+1; 58 b += tmp+1; 59 tmp = calc(a,b); 60 sum += tmp; 61 if(sum >= len2-2){ 62 ans = i; 63 break; 64 } 65 a += tmp+1; 66 b += tmp+1; 67 tmp = calc(a,b); 68 sum += tmp; 69 if(sum >= len2-2){ 70 ans = i; 71 break; 72 } 73 } 74 printf("Case #%d: %d ",cs++,ans); 75 } 76 return 0; 77 }