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, ..., a_n. 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 v_i-th array element equal to x_i. In other words, perform the assignment a_vi = x_i.
2. Increase each array element by y_i. In other words, perform n assignments a_i = a_i + y_i (1 ≤ i ≤ n).
3. Take a piece of paper and write out the q_i-th array element. That is, the element a_qi.

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;
}