uva 12429 Finding Magic Triplets 树状数组

题意:求有几组满足 a+b^2-c^3=0(mod k) 1<=a<=b<=c<=n (a,b,c)。

思路：树状数组

注意到 当b固定时 考虑a 1~b%k有b/k+1  b%k+1~k 有b/k个

从大至小枚举b，每次tree[b^3%k]++  查询 (1+b*b)%k~(b%k+b*b)%k 总和sum*(b/k+1)+(sum总-sum)*b/k

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define NMAX 100001
int n,k;
int tree[NMAX];
long long ans;
int lowbit(int x)
{
    return x&(x^(x-1));
}
void update(int i)
{
    while(i<=k)
    {
        tree[i]++;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int ret=0;
    while(i>0)
    {
        ret+=tree[i];
        i-=lowbit(i);
    }
    return ret;
}
void add(int Min,int Max,int t)
{
    long long ret=0;
    if(Min<Max) ret=sum(Max)-sum(Min);    
    else ret=sum(k)-sum(Min)+sum(Max);
    ans+=t*ret+(t-1)*(sum(k)-ret);
}
int main()
{
    int T;
    int i,j;
    int b;
    int Min,Max;
    int temp;
    
    scanf("%d",&T);
    for(j=1;j<=T;j++)
    {
        ans=0;
        memset(tree,0,sizeof(tree));
        scanf("%d%d",&n,&k);
        for(b=n;b>0;b--)
        {
            temp=((long long)b*b*b-1)%k+1;
            update(temp);
            temp=((long long)b*b-1)%k+1;
            Max=(b%k+temp+k-1)%k+1;
            Min=temp%k;
            add(Min,Max,(b-1)/k+1);
        }
        printf("Case %d: %lld\n",j,ans);
    }
    return 0;
}