Codeforces Round #116 (Div. 2, ACM-ICPC Rules) Letter（DP 枚举）

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

 【题意】给你一个字符串，每一次操作可以将大写字母改成小写，反之亦然。问经过最少的操作数，使得此串的大写字母全部都在小写字母的左边。

【分析】感觉跟DP没啥关系，直接枚举...

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long LL;
const int N = 1e5+50;
const int mod = 1e9+7;
int n,a,b,k;
int dp[N],sum[N];
char str[N];
int main(){
    scanf("%s",str);
    int len=strlen(str);
    for(int i=len-1;i>=0;i--){
        sum[i]=sum[i+1]+(str[i]>='a'?0:1);
    }
    int s=0,ans=sum[0];
    for(int i=0;i<len;i++){
        s+=(str[i]>='a'?1:0);
        ans=min(ans,sum[i+1]+s);
    }
    printf("%d
",ans);
    return 0;
}