HDU1166 敌兵布阵
作者:互联网
此题目是树状数组的模板题,没有任何技术含量,记忆吧~
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 50010;
int n;
// t[i]表示树状数组i结点覆盖的范围和
int t[N];
//返回非负整数x在二进制表示下最低位1及其后面的0构成的数值
int lowbit(int x) {
return x & -x;
}
//将序列中第x个数加上k
void add(int x, int k) {
for (int i = x; i <= n; i += lowbit(i)) t[i] += k;
}
//查询序列前x个数的和
int sum(int x) {
int sum = 0;
for (int i = x; i; i -= lowbit(i)) sum += t[i];
return sum;
}
int main() {
int T;
char op[100]; //读入指令
cin >> T;
for (int k = 1; k <= T; k++) {
scanf("%d", &n);
memset(t, 0, sizeof t);
for (int i = 1; i <= n; i++) {
int x;
scanf("%d", &x);
add(i, x); // i这个营地有x个士兵
}
//输出
cout << "Case " << k << ":" << endl;
while (scanf("%s", op) && *op != 'E') {
int x, y;
scanf("%d%d", &x, &y);
if (*op == 'Q')
cout << sum(y) - sum(x - 1) << endl; //计算区间内士兵个数,前缀和
else if (*op == 'A')
add(x, y); //在x号营地增加y名士兵
else
add(x, -y); //在x号营地减少y名士兵
}
}
return 0;
}
标签:题目,树状,int,long,敌兵,数组,HDU1166,include,布阵 来源: https://www.cnblogs.com/littlehb/p/16221667.html