POJ

题目链接：http://poj.org/problem?id=3468点击打开链接

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 116646 Accepted: 36248Case 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

线段树基础 区间查询区间更新

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;#define maxn 1000000+5struct xjy{    long long int left;    long long int right;    long long int target;    long long int sum;    long long int lazy;    long long int time;};xjy tree[maxn<<2];int a[maxn];long long int ans=0;void build(long long int i,long long int left ,long long int right){    if(left==right)    {        tree[i].left=left;        tree[i].right=right;        tree[i].sum=a[left];        return ;    }    long long int mid=(left+right)>>1;    build(i<<1,left,mid);    build (i<<1|1,mid+1,right);    tree[i].left=left;    tree[i].right=right;    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;}void updateregion(long long int i,long long int left,long long int right,long long int val){   if(tree[i].lazy)    {        if(tree[i].left==tree[i].right)        {            tree[i].lazy=0;            tree[i].target=0;        }        else        {                tree[i<<1].lazy=1;                tree[i<<1].target+=tree[i].target;                tree[i<<1].sum+=tree[i].target*(tree[i<<1].right-tree[i<<1].left+1);                tree[i<<1|1].lazy=1;                tree[i<<1|1].target+=tree[i].target;                tree[i<<1|1].sum+=tree[i].target*(tree[i<<1|1].right-tree[i<<1|1].left+1);                tree[i].lazy=0;                tree[i].target=0;        }    }    if(tree[i].left==left&&tree[i].right==right)    {        tree[i].lazy=1;        tree[i].target+=val;        tree[i].sum+=tree[i].target*(tree[i].right-tree[i].left+1);        return;    }    long long int mid=(tree[i].left+tree[i].right)>>1;    if(right<=mid)        updateregion(i<<1,left,right,val);    else if(left>mid)        updateregion(i<<1|1,left,right,val);    else    {        updateregion(i<<1,left,mid,val);        updateregion(i<<1|1,mid+1,right,val);    }    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;}void queryregion(long long int i,long long int left,long long int right){    if(tree[i].lazy)    {        if(tree[i].left==tree[i].right)        {            tree[i].lazy=0;            tree[i].target=0;        }        else        {                tree[i<<1].lazy=1;                tree[i<<1].target+=tree[i].target;                tree[i<<1].sum+=tree[i].target*(tree[i<<1].right-tree[i<<1].left+1);                tree[i<<1|1].lazy=1;                tree[i<<1|1].target+=tree[i].target;                tree[i<<1|1].sum+=tree[i].target*(tree[i<<1|1].right-tree[i<<1|1].left+1);                tree[i].lazy=0;                tree[i].target=0;        }    }    if(left==tree[i].left&&right==tree[i].right)    {        ans+=tree[i].sum;        return;    }    long long int mid=(tree[i].left+tree[i].right)>>1;    if(right<=mid)        queryregion(i<<1,left,right);    else if(left>mid)        queryregion(i<<1|1,left,right);    else    {        queryregion(i<<1,left,mid);        queryregion(i<<1|1,mid+1,right);    }    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;}int main(){    long long int n,m;    long long int t;    scanf("%lld%lld",&n,&m);    for(int i=1;i<maxn;i++)        {            tree[i].lazy=0;            tree[i].left=0;            tree[i].right=0;            tree[i].target=0;            tree[i].sum=0;            a[i]=0;        }    for(int i=1;i<=n;i++)        scanf("%lld",&a[i]);    build(1,1,n);    for(int i=1;i<=m;i++)        {            string s;            cin >> s;            if(s[0]=='Q')            {                long long int l,r;                scanf("%lld%lld",&l,&r);                ans=0;                queryregion(1,l,r);                printf("%lld\n",ans);            }            else            {                long long int l,r,mid;                scanf("%lld%lld%lld",&l,&r,&mid);                updateregion(1,l,r,mid);            }        }}