hdu1166 敌兵布阵 一维树状数组

Problem:
给了n个数，有三个操作，增加某个值，减少某个值，求某段区间的和。
Solution:
利用线段树或者树状数组都可以，快速求区间和。
notes:
输入输出不加速会超时。

#include<cstdio>#include<iostream>#include<sstream>#include<cstdlib>#include<cmath>#include<cctype>#include<string>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<set>#include<map>#include<ctime>#include<vector>#include<fstream>#include<list>#include <iomanip>using namespace std;typedef long long ll;typedef unsigned long long ull;#define ms(s) memset(s,0,sizeof(s))const double PI = 3.141592653589;const int INF = 0x3fffffff;long long c[100010];int n;inline int lowbit(int x) {    return x & -x;}void update(int idx, int v) {    while (idx <= n) {        c[idx] += v;        idx += lowbit(idx);    }}long long query(int x) {// The sum of 1 to x    long long ans = 0;    while(x > 0) {        ans += c[x];        x -= lowbit(x);    }    return ans;}int main() {    //        freopen("/Users/really/Documents/code/input","r",stdin);    //    freopen("/Users/really/Documents/code/output","w",stdout);    ios_base::sync_with_stdio(false);    cin.tie(0);    int t;    cin >> t;    for(int k = 1; k <= t; k++) {        ms(c);        cout << "Case " << k << ":\n";        cin >> n;        int a;        for(int i = 1; i <= n; i++) {            cin >> a;            update(i, a);        }        string s;        int l, r;        while(cin >> s) {            if(s == "End")                break;            else if(s == "Query") {                cin >> l >> r;                cout << query(r) - query(l-1) << '\n';            }            else if(s == "Add") {                cin >> l >> r;                update(l, r);            }            else if(s == "Sub") {                cin >> l >> r;                update(l, -r);            }        }    }    return 0;}