标签:ace .com 替代 memset using control acm ems ret
推荐博客:https://segmentfault.com/a/1190000003914228
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模板
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 const int maxn = 1e6+10; 6 char str[2*maxn]; 7 int b[2*maxn]; 8 9 int main() { 10 int n; 11 scanf("%d", &n); 12 while(n--) { 13 scanf("%s", str); 14 int len = strlen(str), j = 0, ma = 0; 15 for(int i = len; i >= 0; i--) { 16 str[2*i+2] = str[i]; 17 str[i*2+1] = ‘#‘; 18 } 19 str[0] = ‘@‘; 20 for(int i = 2; i <2*len+1; i++) { 21 if(j+b[j] > i) b[i] = min(b[2*j-i], j+b[j]-i); //i在pos左边 22 else b[i] = 1;//i在pos右边 23 while(str[i-b[i]] == str[i+b[i]]) b[i]++; 24 if(j+b[j] < i+b[i]) j = i;//最长的中心位置 25 if(ma < b[i]) ma = b[i];//回文串的最长长度+1 26 } 27 printf("%d\n", ma-1); 28 } 29 return 0; 30 }
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小刘
1 #include <iostream> 2 #include <string.h> 3 using namespace std; 4 const int maxn = 1e6+10; 5 char str[maxn]; 6 int main(){ 7 int n; 8 cin>>n; 9 while( n -- ){ 10 memset( str, 0, sizeof( str ) ); 11 cin>>str; 12 int len = strlen(str),man = -1,j; 13 for( int i = 0; i < len; i ++ ){ 14 for( j = 0; i - j >= 0 && i + j < len; j ++ ){ 15 if( str[i-j] != str[i+j] ) break; 16 } 17 man = max( man, 2 * j - 1) ; 18 for( j = 0; i - j >= 0 && i + j + 1 < len; j ++ ){ 19 if( str[i - j] != str[i + j + 1] ) break; 20 } 21 man = max( man, 2 * j ); 22 } 23 cout<<man<<endl; 24 } 25 }
输出最大回文子串
1 #include <iostream> 2 #include <ctype.h> 3 #include <stdio.h> 4 #include <string.h> 5 using namespace std; 6 const int maxn = 110000+10; 7 char str[maxn], s[maxn]; 8 int p[maxn]; 9 int main(){ 10 int len, m = 0, man = 0, x, y; 11 int i, j; 12 fgets(str, 1000, stdin); //fgets(s, n, stdin); 从键盘读入长度为不大于n的字符完美替代gets 13 // cout<<s<<endl; 14 len = strlen(str); 15 for(i = 0; i < len; i ++){ 16 if(isalpha(str[i])){ 17 p[m] = i; 18 s[m++] = toupper(str[i]); 19 } 20 } 21 // cout<<s<<endl; 22 // cout<<m<<endl; 23 for(i = 0; i < m; i ++){ 24 for(j = 0; i - j >= 0 && i + j < m; j ++){ 25 if(s[i-j] != s[i+j]) break; 26 if(man < 2*j+1) man = 2*j+1, x = p[i-j], y = p[i+j]; 27 } 28 for(j = 0; i - j >= 0 && i + j + 1 < m; j ++){ 29 if(s[i - j] != s[i + j + 1]) break; 30 if(man < 2*j+2) man = 2*j+2, x = p[i-j], y = p[i+j+1]; 31 } 32 } 33 for(i = x; i <= y; i ++){ 34 printf("%c", str[i]); 35 } 36 printf("\n"); 37 return 0; 38 }
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http://acm.timus.ru/problem.aspx?space=1&num=1297
输出最长回文串
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 const int maxn=1e3+10; 7 char str[2*maxn], ss[maxn]; 8 int b[2*maxn]; 9 10 int main(){ 11 int n; 12 scanf("%s",ss); 13 int len = strlen(ss), j = 0, ma = 0, id; 14 for(int i = len; i >= 0; i--){ 15 str[2*i+2]=ss[i]; 16 str[i*2+1]=‘#‘; 17 } 18 str[0]=‘@‘; 19 for(int i = 2; i < 2*len+1; i++){ 20 if(j+b[j] > i) b[i] = min(b[2*j-i], j+b[j]-i); 21 else b[i] = 1; 22 while(str[i-b[i]] == str[i+b[i]]) b[i]++; 23 if(j+b[j] < i+b[i]) j = i; 24 if(ma < b[i]) ma = b[i], id = j/2; 25 } 26 //printf("%d %d\n", id, ma); 27 ma--; 28 id = id - ma/2 + (ma%2 == 0); 29 ss[id+ma-1] = 0; 30 //printf("%d\n", id); 31 //printf("%d\n",ma); 32 printf("%s\n", ss+id-1); 33 return 0; 34 }
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标签:ace .com 替代 memset using control acm ems ret
原文地址:https://www.cnblogs.com/wenbao/p/5772411.html