POJ 3468 A Simple Problem with Integers（区间更新+区间求和）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 78622 Accepted: 24231Case 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

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




ac代码：

#include<stdio.h>struct s{int left;int right;__int64 sum;__int64 lazy;}tree[1000010];__int64 num[100010];void pushdown(int len,int i){if(tree[i].lazy){tree[i*2].lazy+=tree[i].lazy;tree[i*2+1].lazy+=tree[i].lazy;tree[i*2].sum+=tree[i].lazy*(len-(len/2));tree[i*2+1].sum+=tree[i].lazy*(len/2);tree[i].lazy=0;}}void build(int l,int r,int i){tree[i].left=l;tree[i].right=r;tree[i].lazy=0;if(l==r){tree[i].sum=num[l];}else{int mid=(l+r)/2;build(l,mid,i*2);build(mid+1,r,i*2+1);tree[i].sum=tree[i*2].sum+tree[i*2+1].sum;    }}void update(int l,int r,__int64 c,int i){int mid=(tree[i].right+tree[i].left)/2;if(tree[i].left>=l&&tree[i].right<=r){tree[i].lazy+=c;tree[i].sum+=c*(tree[i].right-tree[i].left+1);return;}pushdown(tree[i].right-tree[i].left+1,i);    if(l<=mid)    {        update(l,r,c,i*2);    }    if(mid<r)    {        update(l,r,c,i*2+1);    }    tree[i].sum=tree[i*2].sum+tree[i*2+1].sum;}__int64 query(int l,int r,int i){int mid=(tree[i].left+tree[i].right)/2;__int64 ans=0;if(tree[i].left>=l&&tree[i].right<=r){return tree[i].sum;}pushdown(tree[i].right-tree[i].left+1,i);if(l<=mid){ans+=query(l,r,i*2);}if(r>mid){ans+=query(l,r,i*2+1);}tree[i].sum=tree[i*2].sum+tree[i*2+1].sum;return ans;}int main(){int i;int a,b,n,m;__int64 d;char ch[10];while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++)scanf("%I64d",&num[i]);build(1,n,1);for(i=0;i<m;i++){scanf("%s",ch);if(ch[0]=='Q'){scanf("%d%d",&a,&b);printf("%I64d\n",query(a,b,1));}else{scanf("%d%d%I64d",&a,&b,&d);update(a,b,d,1);}}}return 0;}