线段树题型3

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 60621		Accepted: 18475
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... ,A_N. 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, ... ,A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... ,A_b.

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

题意：有一些数，然后执行一些操作，C代表把某个区间各个数加上一个数字，Q询问某个区间的总和是多少

代码：

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

struct comp

{

int l,r,mid;

__int64 nsum;//记录该区间内的和

__int64 lnc;//记录该区间加的数的总和

}data[400010];

int num[100010];

void build(int l,int r,int k)

{

data[k].l=l;

data[k].r=r;

data[k].mid=(l+r)/2;

data[k].lnc=0;

if(l==r)

{

data[k].nsum=num[l];//初始值

return ;

}

build(l,data[k].mid,2*k);

build(data[k].mid+1,r,2*k+1);

data[k].nsum=data[2*k].nsum+data[2*k+1].nsum;//把下一层的数加到上一层，更新上一层

}

void updata(int l,int r,__int64 c,int k)

{

if(data[k].l==l&&data[k].r==r)

{

data[k].lnc+=c;

return;

}

data[k].nsum+=(r-l+1)*c;

if(r<=data[k].mid)

updata(l,r,c,2*k);

else if(l>data[k].mid)

updata(l,r,c,2*k+1);

else

{

updata(l,data[k].mid,c,2*k);

updata(data[k].mid+1,r,c,2*k+1);

}

__int64 search(int l,int r,int k)

{

if(data[k].l==l&&data[k].r==r)

{

return data[k].nsum+data[k].lnc*(r-l+1);

}

data[k].nsum+=(data[k].r-data[k].l+1)*data[k].lnc;

updata(data[k].l,data[k].mid,data[k].lnc,2*k);

updata(data[k].mid+1,data[k].r,data[k].lnc,2*k+1);

data[k].lnc=0;

if(r<=data[k].mid)

return search(l,r,2*k);

else if(l>data[k].mid)

return search(l,r,2*k+1);

else

return search(l,data[k].mid,2*k)+search(data[k].mid+1,r,2*k+1);

}

int main()

{

int n,m,i,a,b,c;

char ch;

while(scanf("%d%d",&n,&m)!=EOF)

{

for(i=1;i<=n;i++)

{

scanf("%d",&num[i]);

}

build(1,n,1);

for(i=1;i<=m;i++)

{

getchar();

scanf("%c",&ch);

if(ch=='Q')

{

scanf("%d%d",&a,&b);

printf("%I64d\n",search(a,b,1));

}

if(ch=='C')

{

scanf("%d%d%d",&a,&b,&c);

updata(a,b,c,1);

}

return 0;

}