poj 3468 树状数组

http://kenby.iteye.com/blog/962159

///我只想存个代码,思路来源解法都是上面那个网站看的

#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 100000+10;
long long det[N],num[N],sum[N];
long long  c1[N],c2[N];
int l,r,m,n;
char st[10];
int lowbit(int x)
{
    return x&(-x);
}
void update(long long *array,int l,long long val)
{
    while(l<=n)
    {
        array[l]+=val;
        l+=lowbit(l);
    }
}
long long getsum(long long *array,int l)
{
    long long ans = 0;
    while(l>0)
    {
        ans+=array[l];
        l-=lowbit(l);
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&num[i]);
        sum[i] = sum[i-1]+num[i];
    }
    while(m--)
    {
        scanf("%s",st);
        if(st[0]=='C')
        {
            scanf("%d %d %d",&l,&r,&t);
            update(c1,l,t);
            update(c1,r+1,-t);
            update(c2,l,t*l);
            update(c2,r+1,-t*(r+1));
        }
        else
        {
            scanf("%d %d",&l,&r);
            long long xx = 0;
            xx = sum[r] - sum[l-1];
            xx+= (r+1)*getsum(c1, r) - getsum(c2, r);
            xx -= (l*getsum(c1, l-1) - getsum(c2, l-1));
            printf("%lld\n",xx);
        }
    }
    return 0;
}