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 root
using 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;
}
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 60817 | Accepted: 18545 | |
| 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
POJ3468_A Simple Problem with Integers(线段树/成段更新)
原文：http://www.cnblogs.com/lxjshuju/p/7230594.html