PAT 乙级 1043 输出PATest

解题思路：对于c来说hash函数求出各个符号的数量，不为0时就输出即可，python利用字符串count函数就可以了。
c语言代码如下:

#include<stdio.h>
int output_c(char c,int *count)
{
	if(*count!=0)
	{
		putchar(c);
		(*count)--;
	}
	return 1;
}
int main()
{
	char a[10001];
	int a_count[128]={0};
	int i;
	scanf("%s",a); 
	for(i=0;a[i]!='\0';i++)
	{
		a_count[a[i]]++;
	}
	int count_P=a_count['P'],count_A=a_count['A'],count_T=a_count['T'];
	int count_e=a_count['e'],count_s=a_count['s'],count_t=a_count['t'];
	int count=count_P+count_A+count_T+count_e+count_s+count_t;
	for(i=0;i<count;i++)
	{
	   output_c('P',&count_P);
	   output_c('A',&count_A);
	   output_c('T',&count_T);
	   output_c('e',&count_e);
	   output_c('s',&count_s);
	   output_c('t',&count_t);
	    		
	}
	printf("\n");
	return 0;
}

python代码:

s=input()
out='PATest'
count=0
count_out=[0]*6
for i in out:
    index=out.index(i)
    count_out[index]=s.count(i)
    count+=count_out[index]
for i in range(count):
    for i in out:
        index=out.index(i)
        if count_out[index]>0:
            count_out[index]-=1
            print(i,end='')
print()