一維樹狀數組區間更新區間查詢

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推理過程

維護兩個數組，d[i] = a[i] - a[i - 1]，f[i] = i * d[i]，剩下的，區間更新和區間查詢就和【一維樹狀數組區間更新單點查詢】一樣了。

實現原始碼

#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;}

 

一維樹狀數組區間更新區間查詢