Lightoj 1112

1112 - Curious Robin Hood

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Time Limit: 1 second(s)

Memory Limit: 64 MB

Robin Hood likes to loot rich people since he helps the poor people with this money. Instead of keeping all the money together he does another trick. He keeps nsacks where he keeps this money. The sacks are numbered from 0 to n-1.

Now each time he can he can do one of the three tasks.

1)                  Give all the money of the i^th sack to the poor, leaving the sack empty.

2)                  Add new amount (given in input) in the i^th sack.

3)                  Find the total amount of money from i^th sack to j^th sack.

Since he is not a programmer, he seeks your help.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case contains two integers n (1 ≤ n ≤ 10⁵) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers in the range [0, 1000]. The i^thinteger denotes the initial amount of money in the i^th sack (0 ≤ i < n).

Each of the next q lines contains a task in one of the following form:

1 i        Give all the money of the i^th (0 ≤ i < n) sack to the poor.

2 i v     Add money v (1 ≤ v ≤ 1000) to the i^th (0 ≤ i < n) sack.

3 i j      Find the total amount of money from i^th sack to j^th sack (0 ≤ i ≤ j < n).

Output

For each test case, print the case number first. If the query type is 1, then print the amount of money given to the poor. If the query type is 3, print the total amount from i^th to j^th sack.

Sample Input	Output for Sample Input
1 5 6 3 2 1 4 5 1 4 2 3 4 3 0 3 1 2 3 0 4 1 1	Case 1: 5 14 1 13 2

Notes

Dataset is huge, use faster I/O methods.

题意：给你N个数和M次查询，查询分三种

一，1 i   表示查询i处的值。并把 i 处的值变为0。

二，2 i v 表示将 i 处 值添加 v。

三，3 x y 查询 区间[x, y]的值。

水题。直接用树状数组就能够了，不是必需用线段树。

AC代码：注意下标是从0開始到N-1的

#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define MAXN 100000+10
#define MAXM 60000+10
#define INF 1000000
#define eps 1e-8
using namespace std;
int N, M;
int C[MAXN<<1];
int k = 1;
int lowbit(int x)
{
    return x & (-x);
}
int sum(int x)
{
    int s = 0;
    while(x > 0)
    {
        s += C[x];
        x -= lowbit(x);
    }
    return s;
}
void update(int x, int d)
{
    while(x <= N)
    {
        C[x] += d;
        x += lowbit(x);
    }
}
void solve()
{
    int x, y, d, op;
    printf("Case %d:
", k++);
    while(M--)
    {
        scanf("%d", &op);
        if(op == 1)
        {
            scanf("%d", &x);
            x++;
            printf("%d
", sum(x) - sum(x-1));
            int t = sum(x) - sum(x-1);
            update(x, -t);
        }
        else if(op == 2)
        {
            scanf("%d%d", &x, &d);
            x++;
            update(x, d);
        }
        else
        {
            scanf("%d%d", &x, &y);
            x++, y++;
            printf("%d
", sum(y) - sum(x-1));
        }
    }
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &N, &M);
        memset(C, 0, sizeof(C));
        int a;
        for(int i = 1; i <= N; i++)
        {
            scanf("%d", &a);
            update(i, a);
        }
        solve();
    }
    return 0;
}