Codeforces 180C Letter【dp】

C. Letter

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.

Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero
or more lowercase letters.

To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question:
what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.

Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed
105.

Output
Print a single number — the least number of actions needed to make the message
fancy.

Examples

Input

PRuvetSTAaYA

Output

5

Input

OYPROSTIYAOPECHATALSYAPRIVETSTASYA

Output

0

Input

helloworld

Output

0

题目大意：

给你一个字符串，我们每一次操作都可以将一个大写字母变成任意小写字母，当然同理也可以将小写字母变成任意大写字母，问最少操作多少次，能够使得字符串变成前边都是大写字母，后边都是小写字母。

思路：

1、考虑dp，设定dp【i】【2】，其中dp【i】【0】表示dp到第i位，当前字母变成了小写字母的最小花费，其中dp【i】【1】表示dp到第i位，当前字母变成了大写字母的最小花费。

2、那么不难推出其状态转移方程：

if（当前字母为小写字母）

dp【i】【0】=min（dp【i-1】【0】，dp【i-1】【1】）；

dp【i】【1】=dp【i-1】【1】+1；

else（即当前字母为大写字母）

dp【i】【0】=min（dp【i-1】【0】，dp【i-1】【1】）+1；

dp【i】【1】=dp【i-1】【1】；

其最终解为min（dp【n-1】【0】，dp【n-1】【1】）；

Ac代码：

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
char a[100050];
int dp[100050][2];
int main()
{
while(~scanf("%s",a))
{
int n=strlen(a);
for(int i=0;i<n;i++)
{
if(islower(a[i]))
{
dp[i][0]=min(dp[i-1][0],dp[i-1][1]);
dp[i][1]=dp[i-1][1]+1;
}
else
{
dp[i][0]=min(dp[i-1][0],dp[i-1][1])+1;
dp[i][1]=dp[i-1][1];
}
}
printf("%d\n",min(dp[n-1][0],dp[n-1][1]));
}
}