題目鏈接

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題意：
給定一個序列，有三個操作

• 1 i 輸出第i個數的大小，并將其變為0
• 2 i v 第i個數增加v
• 3 i j 輸出第i個數到第j個數的和

樹狀數組模板題

#include <stdio.h> #include <iostream> #include <string.h> #include <stdlib.h> #include <vector> #include <algorithm> #include <queue> #include <map> #include <stack> #include <string> #include <math.h> #include <bitset> #include <ctype.h> using namespace std; typedef pair<int,int> P; typedef long long LL; const int INF = 0x3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-9; const int N = 1e5 + 5; const int mod = 1e9 + 7; int t, kase = 0; int c[N]; int n,q,type,x,y; inline int lowbit(int i) {return i&(-i); } void add(int x, int val) {while(x <= n){c[x] += val;x += lowbit(x);} } int sum(int x) {int res = 0;while(x){res += c[x];x -= lowbit(x);}return res; }int main() {scanf("%d", &t);while(t--){printf("Case %d:\n", ++kase);scanf("%d%d", &n, &q);memset(c, 0, sizeof(c));for(int i = 1; i <= n; i++){scanf("%d", &x);add(i, x);}while(q--){scanf("%d", &type);if(type == 1){scanf("%d", &x);x++;int ans = sum(x) - sum(x-1);add(x, -ans);printf("%d\n", ans);}else if(type == 2){scanf("%d%d", &x,&y);x++;add(x, y);}else{scanf("%d%d", &x, &y);x++;y++;int ans = sum(y) - sum(x-1);printf("%d\n", ans);}}}return 0; }

轉載于:https://www.cnblogs.com/Alruddy/p/7471061.html

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