http://codeforces.com/gym/101341/problem/B
ProblemB. Pursuing the Happiness
Input file: standard input
Output file: standard output
Time limit: 2 seconds
Memory limit: 256 megabytes
Mike wants to find a substring «happiness» in the string s, but Constantine cannot allow this and decided to hinder him. He is planning to swap two characters on two different positions in the string s so that Mike wouldn’t be able to find what he looks for. Which two characters Constantine should swap?
Input
The only line contains from 2 to 2*10^5 lowercase Latin letters — the string s, in which Mike wants to find a substring «happiness».
Output
If Constantine succeeds in achieving his goal, in the first line output «YES» without quotes. In the second line output two distinct integers separated by a space — the positions of characters in the string s, which Constantine should swap. Positions in the string are numbered from one. If there are several possible answers, output any of them.
If for any choice of Constantine Mike still would be able to find a substring «happiness», in the only line output «NO» without quotes.
Examples
standard input
pursuingthehappiness
standard output
YES
15 18
standard input
happinessformehappinessforyouhappinessforeverybodyfreeandletnoonebeleftbehind
standard output
NO
由于数据量较小，不用ＫＭＰ也可。
1 #include <stdio.h> 2 #include <string.h> 3 4 int main() 5 { 6 int len, i, j, N; 7 char a[200005], b[15] = "happiness", t; 8 int re[5], num; 9 N = 9; 10 num = 0; 11 scanf("%s", a); 12 len = strlen(a); 13 for(i=0; i+N-1<len; i++) 14 { 15 for(j=0; j<N; j++) 16 { 17 if(a[i+j]!=b[j]) break; 18 } 19 if(j==N) 20 { 21 re[num++] = i+1; 22 if(num>=3) break; 23 } 24 } 25 if(len < 9) 26 { 27 printf("YES\n"); 28 printf("%d %d\n", 1, 2); 29 } 30 else if(i+N-1<len) printf("NO\n"); 31 else 32 { 33 if(num==0) 34 { 35 t = a[0]; 36 a[0] = a[1]; 37 a[1] = t; 38 39 for(i=0; i+N-1<len; i++) 40 { 41 for(j=0; j<N; j++) 42 { 43 if(a[i+j]!=b[j]) break; 44 } 45 if(j==N) break; 46 } 47 48 if(i+N-1>=len) 49 { 50 printf("YES\n"); 51 printf("%d %d\n", 1, 2); 52 } 53 54 else 55 { 56 printf("YES\n"); 57 printf("%d %d\n", 1, 3); 58 } 59 60 } 61 else if(num==1) 62 { 63 printf("YES\n"); 64 printf("%d %d\n", re[0], re[0]+1); 65 } 66 else 67 { 68 printf("YES\n"); 69 printf("%d %d\n", re[0], re[1]+1); 70 } 71 } 72 return 0; 73 }