POJ 3468 A Simple Problem with Integers (树状数组) (区间修改+区间查询)
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 114049 | Accepted: 35396 | |
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
区间值 修改, 区间查询;
首先 会超出int 范围 所以用 long long ;
区间修改 与单点修改不同
思路:两个数组C1[x]表示该点元素与左边的差值,C2[x]表示的是x*C[x]
因此: 求区间的和 为 :
1. sum(sum(C[j],j<=i)i<=x)
2. =(x+1)*sum(C[i],i<=x)-sum(i*C[i],i<=x);
区间修改[a,b]:
add(c1,a,num);
add(c1,b+1,-num);
add(c2,a,a*num);
add(c2,b+1,-num*(b+1));
区间求和[a,b]:
ll Get_sum(ll x)
{
return Sum(x,c1)*x-Sum(x,c2);
}
ll Get(ll a,ll b)
{
return Get_sum(b+1)-Get_sum(a);// 闭区间从 b+1 ---a;
}
AC代码:
#include <stdio.h> #include <queue> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> #include <string> typedef long long ll; using namespace std; const int N =1000110; ll c1[N],c2[N]; ll air[N]; ll n; void add(ll tree[],ll k,ll num) { while(k<=n) { tree[k]+=num; k+=k&(-k); } } ll Sum(ll x,ll tree[]) { ll sum=0; while(x>0) { sum+=tree[x]; x-=x&(-x); } return sum; } ll Get_sum(ll x) { return Sum(x,c1)*x-Sum(x,c2); } ll Get(ll a,ll b) { return Get_sum(b+1)-Get_sum(a); } int main() { ll q; ll x; //freopen("input.txt","r",stdin); char str[2]; memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); memset(air,0,sizeof(air)); scanf("%lld %lld",&n,&q); { memset(air,0,sizeof(air)); for(int i=1;i<=n;i++) { scanf("%lld",&x); air[i]=air[i-1]+x; } while(q--) { ll a,b,c; scanf("%s",str); if(str[0]=='C') { scanf("%lld %lld %lld",&a,&b,&c); add(c1,a,c); add(c1,b+1,-c); add(c2,a,a*c); add(c2,b+1,-c*(b+1)); } if(str[0]=='Q') { scanf("%lld %lld",&a,&b); ll ans=Get(a,b); ans+= air[b]-air[a-1]; cout<<ans<<endl; } } } return 0; }