【G - A Simple Problem with Integers 树状数组 】

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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.InputThe 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.OutputYou need to answer all Q commands in order. One answer in a line.Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

HintThe sums may exceed the range of 32-bit integers.

#include<cstdio>#include<cstring>#define NUM 100005typedef long long LL;LL C[2][NUM],n,m;LL summ[NUM];LL Lowbit(int x){return x&-x;}void Plus(LL C[],LL pos,LL num){while(pos<=n){C[pos]+=num;pos+=Lowbit(pos);}}LL Sum(LL C[],LL end){LL sum=0;while(end>0){sum+=C[end];end-=Lowbit(end);}return sum;}int main(){LL cas,Q,l,r,tem,change;while(~scanf("%lld%lld",&cas,&Q)){memset(C,0,sizeof(C));n=cas;for(int i=1;i<=n;i++)scanf("%lld",&summ[i]);summ[0]=0;for(int i=1;i<=n;i++)summ[i] += summ[i-1];char str[3];for(int i=1;i<=Q;i++){scanf("%s",str);if(strcmp(str,"C")==0){scanf("%lld%lld%lld",&l,&r,&change);Plus(C[0],l,change);Plus(C[1],l,l*change);Plus(C[0],r+1,-change);Plus(C[1],r+1,-change*(r+1));}else {scanf("%lld%lld",&l,&r);LL summary=(Sum(C[0],r)*(r+1)-Sum(C[1],4000r))-(Sum(C[0],l-1)*l-Sum(C[1],l-1));summary+=summ[r]-summ[l-1];printf("%lld\n",summary);}}}return 0;}