数字串

题解：有点像找逆序对，因为子串的长度有限制，所以对于每一个s[i]，当它作为尾部时，头部的范围是(i-R+1,i-L+1)；当它作为头部时，尾部的范围是(i+L-1,i+R-1)。修改的时候先减掉该位置原来的数的贡献。

#pragma warning(disable:4996)
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;

const int maxn = 100005;

int n, q, L, R, co[10][maxn];
char s[maxn];

int lowbit(int x) {
    return x & -x;
}

void add(int o, int pos, int x) {
    for (int i = pos; i <= n; i += lowbit(i)) co[o][i] += x;
    return;
}

int sum(int o, int pos) {
    int res = 0;
    for (int i = pos; i >= 1; i -= lowbit(i)) res += co[o][i];
    return res;
}

int main()
{
    while (scanf("%s", s + 1) != EOF) {
        n = strlen(s + 1);
        scanf("%d%d%d", &q, &L, &R);

        ll ans = 0;
        for (int i = 1; i <= n; i++) {
            int o = s[i] - '0';
            add(o, i, 1);
            for (int j = o + 1; j <= 9; j++) {
                ans += sum(j, i - L + 1) - sum(j, i - R);
            }
        }
    
        for (int i = 1; i <= q; i++) {
            int pos, x;
            scanf("%d%d", &pos, &x);
    
            int o = s[pos] - '0';

            for (int j = o + 1; j <= 9; j++) ans -= sum(j, pos - L + 1) - sum(j, pos - R);
            for (int j = o - 1; j >= 0; j--) ans -= sum(j, min(n, pos + R - 1)) - sum(j, min(n, pos + L - 2));
            add(o, pos, -1);
            add(x, pos, 1);
            for (int j = x + 1; j <= 9; j++) ans += sum(j, pos - L + 1) - sum(j, pos - R);
            for (int j = x - 1; j >= 0; j--) ans += sum(j, min(n, pos + R - 1)) - sum(j, min(n, pos + L - 2));

            s[pos] = '0' + x;

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