2 seconds
256 megabytes
standard input
standard output
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
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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[];
int i,flag[],len,num1,num2;
while(gets(a)){
len=strlen(a);
for(i=;i<len;i++)
flag[i]=;
num1=;
for(i=;i<len;i++){
if(a[i]=='S'&&a[i-]=='Y'){
num1++;
flag[i]=;
flag[i-]=;
}
}
num2=;
for(i=;i<len;i++){
if(flag[i]==&&flag[i-]==){
num2++;
flag[i]=; //其实这里去掉更好
flag[i-]=; //楼上+1。。。
if(a[i]=='Y'&&a[i-]=='S')
num2--;
}
}
if(num2>=)
printf("%d\n",num1+);
else
printf("%d\n",num1);
}
return ;
}
修改了一下:
代码2:
#include<stdio.h>
#include<string.h>
int main(){
char a[];
int i,flag[],len,num1,num2;
while(gets(a)){
len=strlen(a);
for(i=;i<len;i++)flag[i]=;
num1=;
memset(flag,,sizeof(flag));
for(i=;a[i]!='\0';i++){
if(a[i]=='S'&&a[i-]=='Y'){
num1++;
flag[i]=;
flag[i-]=;
}
}
num2=;
for(i=;i<len;i++){
if(flag[i]==&&flag[i-]==){
num2++;
flag[i]=;
flag[i-]=;
if(a[i]=='Y'&&a[i-]=='S')num2--;
else break;
}
}
if(num2>=)
printf("%d\n",num1+);
else
printf("%d\n",num1);
}
return ;
}
其实还有一个代码，但是少了个判断条件，想法是一样的，不贴了。。。
然后！！！在做了cf的这个类似题之后发现了新的问题。。。
等会再说吧，要贴cf的代码了。。。
代码1:
#include<stdio.h>
#include<string.h>
int main(){
char a[];
int i,flag[],len,num1,num2;
while(gets(a)){
len=strlen(a);
for(i=;i<len;i++)flag[i]=;
num1=;
memset(flag,,sizeof(flag));
for(i=;a[i]!='\0';i++){
if(a[i]=='K'&&a[i-]=='V'){
num1++;
flag[i]=;
flag[i-]=;
}
}
num2=;
for(i=;i<len;i++){
if(flag[i]==&&flag[i-]==){
num2++; //这里和那个师大的不一样了。
if(a[i]=='V'&&a[i-]=='K')num2--;
else break;
}
}
if(num2>=)
printf("%d\n",num1+);
else
printf("%d\n",num1);
}
return ;
}
VKKVVVKVKVK这组数据如果改成师大的那个字母就wa了，因为flag[]那里错了，(第二个循环里的标记去掉就可以了) 师大的数据水过了。。。
然后！！！cf的这个可以暴力写。
代码2:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cstdlib>
#include<string>
#define eps 0.000000001
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N=;
char str[N];
char b[N];
int main(){
while(gets(str)){
int len=strlen(str);int maxx=;
if(len==){cout<<<<endl;continue;}
for(int i=;i<len-;i++)if(str[i]=='V'&&str[i+]=='K')
maxx++;
for(int i=;i<len;i++){
char c=str[i];int ans=;
if(str[i]=='V')str[i]='K';
else str[i]='V';
for(int j=;j<len-;j++){
if(str[j]=='V'&&str[j+]=='K')ans++;
}
maxx=max(maxx,ans);
str[i]=c;
}
cout<<maxx<<endl;
}
}
一开始有点没太懂
以VKVKVKVVVV这个为例:第一次变成 KKVKVKVVVV，第二次 VVVKVKVVVV，还是3。。。VKKKVKVVVV ，VKVVVKVVVV，3
每次看他有几个Vk，就这样，最后取最大的，到后面的时候maxx=4，ans=4 所以还是4，但是用暴力写师大的就会超时。
暴力写时间复杂度是O(n^2)，那个flag[]的时间复杂度是O(n)，
因为师大的字符串长度是10000，所以暴力过不了，会超时。而cf的这个题是100，所以没超时。。。