xindoo is
always here

codeforces 315 B.Sereja and Array


地址

B. Sereja and Array


time limit per test


1 second


memory limit per test


256 megabytes


input


standard input


output


standard output


Sereja has got an array, consisting of n integers, a1, a2, …, an.
Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:


  1. Make vi-th array
    element equal to xi.
    In other words, perform the assignment avi = xi.

  2. Increase each array element by yi.
    In other words, perform n assignments ai = ai + yi (1 ≤ i ≤ n).

  3. Take a piece of paper and write out the qi-th
    array element. That is, the element aqi.


给N个元素的数组, 有三种操作 1 是把第i个元素变成v,  2是所有元素都加V, 3 询问第i个元素的值。


我用了树状数组,理论上用线段树也可以做,但树状数组明显要好写点,感觉还要比线段树快些。


树状数组原本用来就区间的和,只要稍微改进一下就和更新点,求点的值, 我们如果更新点x为v(当原来点是0事) 我们update(x , v) 和 update(x, -v) , 这样我们求1到x的和是求到的就是x点的值。


//cf 315 B Sereja and Array
//2013-06-13-20.02
#include <stdio.h>
#include <string.h>

const int maxn = 100005;
int a[maxn];
int n;

inline int lowbit(int x)
{
    return x&-x;
}

int update(int x, int v)
{
    while (x <= n+1)
    {
        a[x] += v;
        x += lowbit(x);
    }
    return 0;
}

int getsum(int x)
{
    int sum = 0;
    while (x)
    {
        sum += a[x];
        x -= lowbit(x);
    }
    return sum;
}

int main()
{
    int m;
    while (scanf("%d %d", &n, &m) != EOF)
    {
        memset(a, 0, sizeof(a));
        int t, op, x, v;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &t);
            update(i, t);
            update(i+1, -t);
        }
        while (m--)
        {
            scanf("%d", &op);
            if (op == 1)
            {
                scanf("%d %d", &x, &v);
                int tmp = getsum(x);
                update(x, -tmp);
                update(x+1, tmp);
                update(x, v);
                update(x+1, -v);
            }
            else if (op == 2)
            {
                scanf("%d", &v);
                update(1, v);
            }
            else
            {
                scanf("%d", &x);
                printf("%d\n", getsum(x));
            }
        }
    }
    return 0;
}



打赏
未经允许不得转载:XINDOO » codeforces 315 B.Sereja and Array
分享到: 更多 (0)

评论 抢沙发

xindoo

联系我联系我们

觉得文章有用就打赏一下文章作者

支付宝扫一扫打赏

微信扫一扫打赏