POJ3468_A Simple Problem with Integers(线段树/成段更新)

解题报告

题意：

略

思路：

线段树成段更新，区间求和。

#include <iostream>#include <cstring>#include <cstdio>#define LL long long#define int_now int l,int r,int rootusing namespace std;LL sum[500000],lazy[500000];void push_up(int root,int l,int r) {    sum[root]=sum[root*2]+sum[root*2+1];}void push_down(int rt,int l,int r) {    if(lazy[rt]) {        int m=(r-l+1);        lazy[rt<<1]+=lazy[rt];        lazy[rt<<1|1]+=lazy[rt];        sum[rt<<1]+=lazy[rt]*(m-(m/2));        sum[rt<<1|1]+=lazy[rt]*(m/2);        lazy[rt]=0;    }}void update(int root,int l,int r,int ql,int qr,LL v) {    if(ql>r||qr<l)return;    if(ql<=l&&r<=qr) {        lazy[root]+=v;        sum[root]+=v*(r-l+1);        return ;    }    int mid=(l+r)/2;    push_down(root,l,r);    update(root*2,l,mid,ql,qr,v);    update(root*2+1,mid+1,r,ql,qr,v);    push_up(root,l,r);}LL q_sum(int root,int l,int r,int ql,int qr) {    if(ql>r||qr<l)return 0;    if(ql<=l&&r<=qr)return sum[root];    push_down(root,l,r);    int mid=(l+r)/2;    return q_sum(root*2,l,mid,ql,qr)+q_sum(root*2+1,mid+1,r,ql,qr);}int main() {    int n,q,i,j,ql,qr;    LL a;    scanf("%d%d",&n,&q);    for(i=1; i<=n; i++) {        scanf("%lld",&a);        update(1,1,n,i,i,a);    }    char str[10];    for(i=1; i<=q; i++) {        scanf("%s",str);        if(str[0]=='Q') {            scanf("%d%d",&ql,&qr);            printf("%lld\n",q_sum(1,1,n,ql,qr));        } else {            scanf("%d%d%lld",&ql,&qr,&a);            update(1,1,n,ql,qr,a);        }    }    return 0;}

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 60817 Accepted: 18545Case 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.