一个简单的问题 `2`

AcWing-243-一个简单的整数问题2open in new window

分析

见《进阶指南》第207页。

$ i = 1 \sum x a i $

$= a 1 + a 2 + . . . + a x $

$= d 1 + (d 1 + d 2 ) + . . . + (d 1 + d 2 + . . . + d x )$

$= x \times d 1 + (x - 1) \times d 2 + . . . + 1 \times d x $

$= (x + 1) \times (d 1 + d 2 + . . . + d x ) - (1 \times d 1 + 2 \times d 2 + . . . + x \times d x )$

$= (x + 1) \times i = 1 \sum x d i - i = 1 \sum x i \times d i $

实现

#include <iostream>
using namespace std;

typedef long long LL;
const int N = 100010;
int n, a[N], m;
LL c[2][N], sum_a[N];

int lowbit (int x) {
    return x & -x;
}
void add (int k, int idx, int val) {
    while (idx <= n) {
        c[k][idx] += val;
        idx += lowbit(idx);
    }
}
LL sum (int k, int idx) {
    LL res = 0;
    while (idx >= 1) {
        res += c[k][idx];
        idx -= lowbit(idx);
    }
    return res;
}

int main () {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++ i) scanf("%d", &a[i]);

    for (int i = 1; i <= n; ++ i) sum_a[i] = sum_a[i - 1] + a[i];
    while (m --) {
        char opt[5];
        scanf("%s", opt);
        
        if (opt[0] == 'C') {
            int l, r, d;
            scanf("%d%d%d", &l, &r, &d);
            add(0, l, d);
            add(0, r + 1, -d);
            add(1, l, l * d);
            add(1, r + 1, -(r + 1) * d);
        } else {
            int l, r;
            scanf("%d%d", &l, &r);
            LL res = sum_a[r] + (r + 1) * sum(0, r) - sum(1, r)
                - (sum_a[l - 1] + l * sum(0, l - 1) - sum(1, l - 1));
            printf("%lld\n", res);
        }
    }
    return 0;
}

# 一个简单的问题 2

# 分析

# 实现

一个简单的问题 `2`

分析

实现