2019CCPC湖南全国邀请赛（广东省赛、江苏省赛） E Hello xtCpc_the input file contains an integer show the maximu-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43497140/article/details/90382867

博客围绕在给定的大小写字母字符串中查找尽可能多的“gdCpc”序列展开，同一位置字母只能用一次。介绍了输入输出格式，给出示例，还提及解题思路为前缀和，将其分成 c,xt,xtC,xtCp,xtCpc 。

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Problem Description

You have a string of lowercase letters.You need to find as many sequence “gdCpc” as possible.But letters in the same position can only be used once。

Input

The input file contains two lines.

The first line is an integer n show the length of string.(1≤n≤2×105)

The second line is a string of length n consisting of lowercase letters and uppercase letters.

Output

The input file contains an integer show the maximum number of different subsequences found.

Sample Input

10 gdCgdCpcpc

Sample Output

思路：前缀和，分成 c,xt,xtC,xtCp,xtCpc

#include<cstdio>
#include<cstring>
using namespace std;
char a[205000];
int falg=1;//xtCpc

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",a);
        /*for(int i=0;i<40000;i++) a[i]='x';
        for(int i=40000;i<80000;i++) a[i]='t';
        for(int i=80000;i<120000;i++) a[i]='C';
        for(int i=120000;i<160000;i++) a[i]='p';
        for(int i=160000;i<200000;i++) a[i]='c';
        a[200000]='\0';*/
        int sum[10]={0},s=0;
        for(int i=0;a[i]!='\0';i++)
        {
            if(a[i]=='x') sum[1]++;
            else if(a[i]=='t')
            {
                if(sum[1]>=1)
                {
                    sum[1]--;
                    sum[2]++;
                }
            }
            else if(a[i]=='C') {
                if(sum[2]>=1)
                {
                    sum[2]--;
                    sum[3]++;
                }
            }
            else if(a[i]=='p') {
                if(sum[3]>=1)
                {
                    sum[3]--;
                    sum[4]++;
                }
            }
            else if(a[i]=='c')
            {
                if(sum[4]>=1)
                {
                    sum[4]--;
                    sum[5]++;
                }
            }
        }
        //scanf("%s",a);
        printf("%d\n",sum[5]);
    }
}