| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 71540 | Accepted: 22049 | |
| 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.
该题的关键是要高效的实现对一个区间的每个数都加上一个数，所以增加了一个数组data表示这一次加的那个数,datb表示去掉这个数的区间和。然后在sum函数里更新这两个值。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int DAT_SIZE = (1<<18)-1 ;
const int MAX_N = 100005;
const int MAX_Q = 100005;
int n,q,a,b,c;
char s[10];
ll data[DAT_SIZE],datb[DAT_SIZE];
void add(int a,int b,int x,int k,int l,int r){
if(a<=l&&r<=b) data[k] += x;
else if(l<b&&a<r) {
datb[k] += (min(b,r)-max(a,l))*x;
add(a,b,x,k*2+1,l,(l+r)/2);
add(a,b,x,k*2+2,(l+r)/2,r);
}
}
ll sum(int a,int b,int k,int l,int r){
if(b<=l||r<=a) return 0;
else if(a<=l&&r<=b) return data[k] * (r-l) + datb[k];
else {
ll res = (min(b,r) - max(a,l)) * data[k];
res += sum(a,b,k*2+1,l,(l+r)/2);
res += sum(a,b,k*2+2,(l+r)/2,r);
return res;
}
}
int main(){
scanf("%d%d",&n,&q);
for(int i=0;i<n;i++) {
scanf("%d",&a);
add(i,i+1,a,0,0,n);
}
while(q--) {
scanf("%s",s);
if(s[0]=='C') {
scanf("%d%d%d",&a,&b,&c);
add(a-1,b,c,0,0,n);
} else {
scanf("%d%d",&a,&b);
printf("%lld\n",sum(a-1,b,0,0,n));
}
}
return 0;
}
A Simple Problem with Integers（POJ-3468）（线段树）
原文地址：http://blog.csdn.net/weizhuwyzc000/article/details/45599417
