poj3468树状数组之区间更新+区间询问

转载来自：CSDN acm_xyyh

Language:  Default A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 54069   Accepted: 16249 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

 /*分析：由于本题更新的时候是区间更新
 所以不能直接去一个个更新区间内的点，肯定会超时
 对于每次更新C(a,b,d)表示区间[a,b]内的值增加d
 用ans[a]表示a~n区间元素增加的值,所以对于C(a,b,d)有:ans[a]+=d,ans[b+1]-=d;
 则每次询问的时候Q(a,b),求a~b的和SUM=sum(a,b)+ans[a]*(b-a+1)+ans[a+1]*(b-a)...+ans[b]//sum(a,b)表示a,b的和
 Sum=sum(a,b)+sum(ans[a+t]*(b-a-t+1))=sum(a,b)+sum(ans[i]*(b-i+1));a<=i<=b;
 Sum=sum(a,b)+(b+1)*sum(ans[i])-sum(ans[i]*i);//1~b所以(b+1)*sum(ans[i]),1~a-1则a*sum(ans[i])
 所以可以用两个树状数组分别表示ans[i]的前缀和 和 ans[i]*i的前缀和
 */
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <algorithm>
 #include <map>
 #include <cmath>
 #include <iomanip>
 #define INF 99999999
 typedef long long LL;
 using namespace std;

 const int MAX=100000+10;
 LL n,q;
 LL sum[MAX],c1[MAX],c2[MAX];

 LL lowbit(LL x){
     return x&(-x);
 }

 void Update(LL x,LL d,LL *c){
     while(x<=n){
         c[x]+=d;
         x+=lowbit(x);
     }
 }

 LL Query(LL x,LL *c){
     LL sum=0;
     while(x>0){
         sum+=c[x];
         x-=lowbit(x);
     }
     return sum;
 }

 int main(){
     char op[3];
     LL x,y,d;
     while(~scanf("%lld%lld",&n,&q)){
         memset(c1,0,sizeof c1);
         memset(c2,0,sizeof c2);
         for(int i=1;i<=n;++i)scanf("%lld",sum+i),sum[i]+=sum[i-1];
         for(int i=0;i<q;++i){
             scanf("%s",op);
             if(op[0] == 'C'){//ans[x]+=d,ans[y+1]-=d
                 scanf("%lld%lld%lld",&x,&y,&d);
                 Update(x,d,c1);
                 Update(y+1,-d,c1);
                 Update(x,x*d,c2);
                 Update(y+1,-(y+1)*d,c2);
             }else{
                 scanf("%lld%lld",&x,&y);
                 printf("%lld\n",sum[y]-sum[x-1]+Query(y,c1)*(y+1)-Query(x-1,c1)*x-Query(y,c2)+Query(x-1,c2));
             }
         }
     }
     return 0;
 }