一维树状数组区间更新区间查询

Posted 2020-10-10

推理过程

维护两个数组，d[i] = a[i] - a[i - 1]，f[i] = i * d[i]，剩下的，区间更新和区间查询就和【一维树状数组区间更新单点查询】一样了。

实现源代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n = 20;
int a[40], d[40], f[40];

int query(int* arr, int x) {
    int res = 0;
    for(int i = x; i > 0; i -= (i & -i))res += arr[i];
    return res;
}

void update(int* arr, int x, int val) {
    for(int i = x; i <= n; i += (i & -i))arr[i] += val;
}

inline int getsum(int x){
    return (x + 1) * query(d, x) - query(f, x);
}

int main() {
    memset(a, 0, sizeof(a));
    memset(d, 0, sizeof(d));
    memset(f, 0, sizeof(f));
    for(int i = 1; i <= n; ++i) {
        a[i] = i;
        int t = a[i] - a[i - 1];
        update(d, i, t);
        update(f, i, t * i);
    }
    for(int i = 1; i <= n; ++i) {
        printf("sum(%d) = %d\\n", i, getsum(i));
    }
    puts("------------------------------------------");
    int x = 1, y = 10, val = 20;
    update(d, x, val);
    update(d, y + 1, -val);
    update(f, x, val * x);
    update(f, y + 1, -val * (y + 1));
    printf("sum(10) = %d\\n", getsum(10));
    return 0;
}