Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string s with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i. e. a letter "K" right after a letter "V") in the resulting string.
The first line will contain a string s consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
VK
1
VV
1
V
0
VKKKKKKKKKVVVVVVVVVK
3
KVKV
1
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
这个题和师大的新生赛的一个题好像好像。。。
Jxnu Group Programming Ladder Tournament 2017
L1-2 叶神的字符串
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 10
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
Problem Description
Input
每组测试数据输入一串由'Y','S'组成的字符串。(字符串长度最多为10000)
Output
Sample Input
YYYS
Sample Output
2
唉,真是让人无语。首先是这个师大的。
代码1:
#include<stdio.h> #include<string.h> int main(){ char a[10050]; int i,flag[10050],len,num1,num2; while(gets(a)){ len=strlen(a); for(i=0;i<len;i++) flag[i]=0; num1=0; for(i=1;i<len;i++){ if(a[i]=='S'&&a[i-1]=='Y'){ num1++; flag[i]=1; flag[i-1]=1; } } num2=0; for(i=1;i<len;i++){ if(flag[i]==0&&flag[i-1]==0){ num2++; flag[i]=1; //其实这里去掉更好 flag[i-1]=1; //楼上+1。。。 if(a[i]=='Y'&&a[i-1]=='S') num2--; } } if(num2>=1) printf("%d ",num1+1); else printf("%d ",num1); } return 0; }
修改了一下:
代码2:
#include<stdio.h> #include<string.h> int main(){ char a[10050]; int i,flag[10050],len,num1,num2; while(gets(a)){ len=strlen(a); for(i=0;i<len;i++)flag[i]=0; num1=0; memset(flag,0,sizeof(flag)); for(i=1;a[i]!='