| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 97217 | Accepted: 30358 | |
| 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
Source
但这个题目求的是某一区间的数组和，而且要支持批量更新某一区间内元素的值，怎么办呢？实际上，
还是可以把问题转化为求数组的前缀和。
首先，看更新操作update(s, t, d)把区间A[s]...A[t]都增加d，我们引入一个数组delta[i]，表示
A[i]...A[n]的共同增量，n是数组的大小。那么update操作可以转化为：
1）令delta[s] = delta[s] + d，表示将A[s]...A[n]同时增加d，但这样A[t+1]...A[n]就多加了d，所以
2）再令delta[t+1] = delta[t+1] - d，表示将A[t+1]...A[n]同时减d
然后来看查询操作query(s, t)，求A[s]...A[t]的区间和，转化为求前缀和，设sum[i] = A[1]+...+A[i]，则
A[s]+...+A[t] = sum[t] - sum[s-1]，
那么前缀和sum[x]又如何求呢？它由两部分组成，一是数组的原始和，二是该区间内的累计增量和, 把数组A的原始
值保存在数组org中，并且delta[i]对sum[x]的贡献值为delta[i]*(x+1-i)，那么
sum[x] = org[1]+...+org[x] + delta[1]*x + delta[2]*(x-1) + delta[3]*(x-2)+...+delta[x]*1
= org[1]+...+org[x] + segma(delta[i]*(x+1-i))
= segma(org[i]) + (x+1)*segma(delta[i]) - segma(delta[i]*i)，1 <= i <= x
这其实就是三个数组org[i], delta[i]和delta[i]*i的前缀和，org[i]的前缀和保持不变，事先就可以求出来，delta[i]和
delta[i]*i的前缀和是不断变化的，可以用两个树状数组来维护。
#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 100010
#define ll long long
using namespace std;
ll c1[N];//c1[i]表示i~n共同增加c1[i]
ll c2[N];//c2[i]表示i~n一共增加c1[i]*i=c2[i]
ll ans[N];//存放的前缀和
ll n,m;
string op;
ll lowbit(ll x)
{
return x&(-x);
}
void update(ll x,ll val,ll *c)
{
while(x<=n)
{
c[x]+=val;
x+=lowbit(x);
}
}
ll getsum(ll x,ll *c)
{
ll s=;
while(x>)
{
s+=c[x];
x-=lowbit(x);
}
return s;
}
int main()
{
freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
while(scanf("%lld%lld",&n,&m)!=EOF)
{
memset(c1,,sizeof c1);
memset(c2,,sizeof c2);
memset(ans,,sizeof ans);
for(int i=;i<=n;i++)
{
scanf("%lld",&ans[i]);
ans[i]+=ans[i-];
}
getchar();
for(int i=;i<=m;i++)
{
cin>>op;
if(op=="C")
{
ll s1,s2,s3;
scanf("%lld%lld%lld",&s1,&s2,&s3);
update(s1,s3,c1);//c1~n共同增加了s3
update(s2+,-s3,c1);//上一步操作使得s2~n多增加了s3所以这一步要减去
update(s1,s1*s3,c2);
update(s2+,-(s2+)*s3,c2);
}
else if(op=="Q")
{
ll s1,s2;
scanf("%lld%lld",&s1,&s2);
ll cur=ans[s2]-ans[s1-];//首先等于s1~s2这个区间的基础值
cur+=getsum(s2,c1)*(s2+)-getsum(s2,c2);//0~s2对前缀和的影响
cur-=getsum(s1-,c1)*(s1)-getsum(s1-,c2);//0~s1对前缀和的影响
printf("%lld\n",cur);
}
}
// for(int i=1;i<=n;i++)
// cout<<getsum(i)<<" ";
// cout<<endl;
}
}