线段树：POJ3468-A Simple Problem with Integers（线段树注意事项）

A Simple Problem with Integers

Time Limit: 10000MS

Memory Limit: 65536K

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

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

4

55

9

15

Hint

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

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解题心得：

1. 这是一个最简单的线段树，题意就是给你一系列数，每次询问l到r的和或者每次在l到r之间的每一个数加上一个数。很简单啊，但是万万没想到比赛居然崩在这个题上面，将r和R传参的时候写反了，tm样例和所有自己造的数据都过了，哎，已砍手。google翻译还吧这题翻译成了神题，我去。
2. 这个题还是很有教训的，线段树的代码比较麻烦，要不停的向上维护，向下传递，不停的传递参数，所以在写的时候一定要注意，写慢一点，不然线段树出现了bug很难找，思路不复杂别死在了手贱上面。线段树的lazy标记的时候一定要记得下移，以及在每次递归之后记得向上维护，向上维护的时候一定要注意父节点和两个子节点的关系，该传递一些什么，别和lazy下移的时候搞混了。

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#include<cstring>
#include<stdio.h>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<math.h>
using namespace std;
const int maxn = 1e6+100;
struct node
{
    long long l,r,sum,lazy;
}tree[maxn<<2];

void pushup(long long root)
{
    tree[root].sum = tree[root<<1].sum + tree[root<<1|1].sum;
}

void pushdown(long long root)
{
    if(tree[root].lazy == 0)
        return ;
    tree[root<<1].lazy += tree[root].lazy;
    tree[root<<1|1].lazy += tree[root].lazy;
    tree[root<<1].sum += (tree[root<<1].r - tree[root<<1].l + 1)*tree[root].lazy;
    tree[root<<1|1].sum += (tree[root<<1|1].r - tree[root<<1|1].l + 1)*tree[root].lazy;
    tree[root].lazy = 0;
}

void buildtree(long long l,long long r,long long root)
{
    tree[root].l = l;
    tree[root].r = r;
    tree[root].lazy = 0;
    if(l == r)
    {
        scanf("%lld",&tree[root].sum);
        return ;
    }
    long long mid = (l + r) >> 1;
    buildtree(l,mid,root<<1);
    buildtree(mid+1,r,root<<1|1);
    pushup(root);
}

long long query(long long L,long long R,long long l,long long r,long long root)
{
    if(l == L && R ==r)
    {
        return tree[root].sum;
    }
    long long mid = (l + r)>>1;
    pushdown(root);
    if(R <= mid)
    {
        return query(L,R,l,mid,root<<1);
    }
    else if(L > mid)
    {
        return query(L,R,mid+1,r,root<<1|1);
    }
    else
        return query(mid+1,R,mid+1,r,root<<1|1)+ query(L,mid,l,mid,root<<1);//这里手贱找了一个小时的bug
    pushup(root);
}

void add(long long L,long long R,long long l,long long r,long long root,long long h)
{
    if(l == L && R ==r)
    {
        tree[root].lazy += h;
        tree[root].sum += (r - l +1)*h;
        return;
    }
    pushdown(root);
    long long mid = (l + r) >> 1;
    if(R <= mid)
        add(L,R,l,mid,root<<1,h);
    else if(L > mid)
        add(L,R,mid+1,r,root<<1|1,h);
    else
    {
        add(L,mid,l,mid,root<<1,h);
        add(mid+1,R,mid+1,r,root<<1|1,h);
    }
    pushup(root);
}

int main()
{
    long long n,m;
    while(scanf("%lld%lld",&n,&m) != EOF)
    {
        long long sum = 0;
        buildtree(1,n,1);
        while(m--)
        {
            long long a,b,h;
            char c[10];//尽量别输入%c,容易挂掉，%s挺好的
            scanf("%s",&c);//这里也要注意一下输入的问题，不同的字符对应的不同个数的int输入
            if(c[0] == 'C')
            {
                long long h;
                scanf("%lld%lld%lld",&a,&b,&h);
                add(a,b,1,n,1,h);
            }
            else if(c[0] == 'Q')
            {
                scanf("%lld%lld",&a,&b);
                sum = query(a,b,1,n,1);
                printf("%lld\n",sum);
            }
        }
    }
    return 0;
}