C 解法, 执行用时: 3ms, 内存消耗: 484KB, 提交时间: 2019-05-13

#include<stdio.h>
#include<string.h>
int main()
{
	char s[100000];
	char t[100000];
	scanf("%s%s",s,t);
	long long lens=strlen(s);
	long long lent=strlen(t);
	long long  numa, numb, total, pos;
	numa=numb=total=pos=0;

	int i,j;

	for (i=0; i<lens; ++i)
	{
        if (i<lent)
		{
            if (t[i] == 'a') 
				numa++;
            else 
				numb++;
        }
        if (s[i] == 'a') 
			total += numb;
        else 
			total += numa;
        if (i>=lens - lent)
		{
            if (t[pos++] == 'a')
				numa--;
            else 
				numb--;
        }
    }
	printf("%lld\n",total);
	return 0;
}

C 解法, 执行用时: 3ms, 内存消耗: 556KB, 提交时间: 2020-08-14

#include<stdio.h>
#include<string.h>
int main()
{
    char s[100000];
    char t[100000];
    scanf("%s%s",s,t);
    long long lens=strlen(s);
    long long lent=strlen(t);
    long long  numa, numb, total, pos;
    numa=numb=total=pos=0;
 
    int i,j;
 
    for (i=0; i<lens; ++i)
    {
        if (i<lent)
        {
            if (t[i] == 'a')
                numa++;
            else
                numb++;
        }
        if (s[i] == 'a')
            total += numb;
        else
            total += numa;
        if (i>=lens - lent)
        {
            if (t[pos++] == 'a')
                numa--;
            else
                numb--;
        }
    }
    printf("%lld\n",total);
    return 0;
}