蓝桥杯 子串分值

问题描述

输入格式

输出格式

输出一个整数表示答案。

样例输入

ababc

样例输出

21

评测用例规模与约定

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<string>
#include<sstream>
#define pr(x) cout << #x << ": " << x << "   "
#define prln(x) cout << #x << ": " << x << endl
using namespace std;
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
int dr[] = {0, 0, -1, 1};
int dc[] = {1, -1, 0, 0};
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
int lc[26]; //某字母在字符串中左边最近一次出现的位置
int rc[26];
int r[MAXN]; //某字母在字符串中右边最近一次出现的位置
char s[MAXN];
int main(){
    scanf("%s", s);
    int len = strlen(s);
    for(int i = 0; i < 26; ++i){
        lc[i] = -1;
        rc[i] = len;
    }
    for(int i = 0; i < len; ++i){
        r[i] = len;
    }

    for(int i = len - 1; i >= 0; --i){
        r[i] = rc[s[i] - 'a'];
        rc[s[i] - 'a'] = i;
    }
    LL ans = 0;
    for(int i = 0; i < len; ++i){
        ans += (LL)(i - lc[s[i] - 'a']) * (r[i] - i);
        lc[s[i] - 'a'] = i;

    }
    printf("%lld\n", ans);
    return 0;
}