POJ 3468 - A Simple Problem with Integers(线段树区间更新)

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 100889 Accepted: 31452
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

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.

题意：
并查集区间更新查询。

解题思路：
模板。

AC代码：

#include<stdio.h>#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int maxn = 1e5+5;long long add[maxn<<2];long long sum[maxn<<2];void PushUp(int rt) {sum[rt] = sum[rt<<1]+sum[rt<<1|1];}void PushDown(int rt,int m){    if(add[rt])    {        add[rt<<1] += add[rt];        add[rt<<1|1] += add[rt];        sum[rt<<1] += add[rt]*(m-(m>>1));        sum[rt<<1|1] += add[rt]*(m>>1);        add[rt] = 0;    }}void Update(int L,int R,int c,int l,int r,int rt){    if(L <= l && R >= r)    {        add[rt] += c;        sum[rt] += c*(r-l+1);        return ;    }    PushDown(rt,r-l+1);    int mid = (l+r)>>1;    if(L <= mid)  Update(L,R,c,lson);    if(R > mid)   Update(L,R,c,rson);    PushUp(rt);}long long query(int L,int R,int l,int r,int rt){    if(L <= l && R >= r)    return sum[rt];    PushDown(rt,r-l+1);    int mid = (l+r)>>1;    long long res = 0;    if(L <= mid)  res += query(L,R,lson);    if(R > mid)   res += query(L,R,rson);    return res;}void build(int l,int r,int rt){    add[rt] = 0;    if(l == r)    {        scanf("%lld",&sum[rt]);        return ;    }    int mid = (l+r)>>1;    build(lson);    build(rson);    PushUp(rt);}int main(){    int n,T;    scanf("%d%d",&n,&T);    build(1,n,1);    while(T--)    {        char op[2];        scanf("%s",&op);        if(op[0] == 'Q')        {            int L,R;            scanf("%d%d",&L,&R);            printf("%lld\n",query(L,R,1,n,1));        }        else        {            int L,R,c;            scanf("%d%d%d",&L,&R,&c);            Update(L,R,c,1,n,1);        }    }    return 0;}