标签：+= init ios values data eal bsp sim void
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
#include<cstring> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; #define maxn 100004 ll a[maxn],c1[maxn],c2[maxn]; ll n,m; ll lowbit(ll x){ return x&(-x); } void updata(ll index,ll x,ll val) { if(index==0) for(int i=x;i<=n;i+=lowbit(i)) c1[i]+=val;//建立树状数 else for(int i=x;i<=n;i+=lowbit(i)) c2[i]+=(x-1)*val;//区间改变 } ll getsum(ll x) { ll sum1=0,sum2=0; for(int i=x;i>0;i-=lowbit(i))//无论变不变化都可用来求和 sum1+=c1[i],sum2+=c2[i];//区间求和//（l,r)的和为getsum(r)-getsum(l-1) return x*sum1-sum2; } int main() { a[0]=0; scanf("%lld%lld",&n,&m); for(int i = 1; i <= n; i++){ scanf("%lld",&a[i]); ll t=a[i]-a[i-1]; updata(0,i,t); updata(1,i,t); //输入初值的时候，也相当于更新了值 } char v; ll ans1,ans2; ll l,r,add; for(int i=1;i<=m;i++){ getchar(); scanf("%c",&v); if(v==‘C‘){ scanf("%lld%lld%lld",&l,&r,&add); updata(0,l,add); updata(0,r+1,-add); updata(1,l,add); updata(1,r+1,-add); } else{ scanf("%lld%lld",&l,&r); ans1=getsum(r); ans2=getsum(l-1); printf("%lld\n",ans1-ans2); } } return 0; }
A Simple Problem with Integers（树状数组区间变化和区间求和）
标签：+= init ios values data eal bsp sim void
原文地址：https://www.cnblogs.com/lipu123/p/12194480.html
