Simple Problem with Integers（POJ 3486）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 137519		Accepted: 42602
Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000. The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000. Each of the next Q lines represents an operation. "C a b c" means adding c to each of Aa, Aa 1, ... , Ab. -10000 ≤ c ≤ 10000. "Q a b" means querying the sum of Aa, Aa 1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

代码语言：javascript复制

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

代码语言：javascript复制

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

题解：又是一道裸地线段树操作。区间查询和区间修改。注意long long。

代码语言：javascript复制

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<map>
#include<cmath>
#include<string>
using namespace std;
typedef long long ll;
ll ans;
struct node
{
    ll l, r, w;
    ll f;
};
struct node tree[100000 * 4   1];
void BuildSegmentTree(int k, int l, int r)
{
    tree[k].l = l;
    tree[k].r = r;
    if(l == r )
    {
        scanf("%lld", &tree[k].w);
        return ;
    }
    int m = (tree[k].l   tree[k].r) >> 1;
    BuildSegmentTree(k << 1, l, m);
    BuildSegmentTree(k << 1 | 1, m   1, r);
    tree[k].w = tree[2 * k].w   tree[2 * k   1].w;
}
void down(int k)
{
    tree[k << 1].f  = tree[k].f;
    tree[k << 1 | 1].f  = tree[k].f;
    tree[k << 1]. w  = tree[k].f * (tree[k * 2].r - tree[k * 2].l   1);
    tree[k << 1 |1].w  = tree[k].f * (tree[k << 1| 1].r - tree[k << 1| 1].l   1);
    tree[k].f = 0;
}
void Lazysum(int k, int x, int y)
{
    if(tree[k].l >= x && tree[k].r <= y)
    {
        ans  = tree[k].w;
        return ;
    }
    if(tree[k].f) down(k);
    int m = (tree[k].l   tree[k].r) / 2;
    if(x <= m) Lazysum(k << 1, x, y);
    if(y > m) Lazysum(k << 1 | 1, x, y);
}
void LazyAdd(int k, int x, int y, int z)
{
    if(tree[k].l >= x && tree[k].r <= y)
    {
        tree[k].w  = z * (tree[k].r - tree[k].l   1);
        tree[k].f  = z;
        return ;
    }
    if(tree[k].f) down(k);
    int m = (tree[k].l   tree[k].r) / 2;
    if(x <= m) LazyAdd(2 *k, x, y, z);
    if(y > m) LazyAdd(2 * k   1, x, y, z);
    tree[k].w = tree[k << 1].w   tree[k << 1 |1].w;
}
char op;
int main()
{
    int N, Q;
    while(~scanf("%d %d",&N, &Q))
    {
        BuildSegmentTree(1,1,N);
        while(Q--)
        {
            int x, y,z;
            getchar();
            scanf("%c %d %d", &op, &x, &y);
            if(op == 'Q')
            {
                ans = 0;
                Lazysum(1,x,y);
                printf("%lldn",ans);
            }
            else if(op == 'C')
            {
                scanf("%d", &z);
                LazyAdd(1,x,y,z);
            }
        }
    }
    return 0;
}

input limit line numbers sum

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