HDU-1166 敌兵布阵 (树状数组模板题——单点更新,区间求和)

题目链接

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define mp make_pair
#define pi acos(-1)
#define pii pair<int, int>
#define pll pair<long long , long long>
#define ll long long
#define ld long double
#define MEMS(x) memset(x, -1, sizeof(x))
#define MEM(x) memset(x, 0, sizeof(x))
#define lowbit(x) x&(-x)
const int inf = 0x3f3f3f3f;
const int maxn = 50001;
using namespace std;
int T, N, c[maxn];
char order[maxn];
void update(int x, int y)
{
    for(int i = x; i <= N; i += lowbit(i))
        c[i] += y;
}
int getsum(int x)
{
    int ans = 0;
    while(x)
    {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}
int main()
{
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    scanf("%d", &T);
    for(int t = 1; t <= T; t++)
    {
        scanf("%d", &N);
        memset(c, 0, sizeof(c));
        for(int i = 1; i <= N; i++)
        {
            int a;
            scanf("%d", &a);
            update(i, a);
        }
        printf("Case %d:\n", t);
        while(scanf("%s", order) != EOF)
        {
            if(!strcmp(order, "End"))
                break;
            if(!strcmp(order, "Add"))
            {
                int i, j;
                scanf("%d %d", &i, &j);
                update(i, j);
            }
            else if(!strcmp(order, "Sub"))
            {
                int i, j;
                scanf("%d %d", &i, &j);
                update(i, -j);
            }
            else if(!strcmp(order, "Query"))
            {
                int i, j;
                scanf("%d %d", &i, &j);
                printf("%d\n", getsum(j) - getsum(i - 1));
            }
        }
    }

}

posted @ 2020-02-15 23:24  DIY-Z  阅读(156)  评论(0编辑  收藏  举报