CF Gym 100463A （树状数组求逆序数）

题意：给你一个序列，和标准序列连线，求交点数。

题解：就是求逆序对个数，用树状数组优化就行了。具体过程就是按照顺序往树状数组了插点（根据点的大小），因为第i大的点应该排在第i位，插进去的时候他前面本该是有i-1个点的，如果少了，那么一定就和这个元素构成了逆序对。

 

#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cstring>

using namespace std;

//#define local
typedef long long  ll;
const int maxn = 1000000+100;
#define lowbit(x) ((x)&-(x))
ll C[maxn];
ll n,a,b;
void add(ll x,ll d)
{
    while(x <= n) {
        C[x]+=d; x+=lowbit(x);
    }
}

ll sum(ll x)
{
    ll ret = 0;
    while(x > 0){
        ret += C[x]; x-=lowbit(x);
    }
    return ret;
}

ll Count()
{
    ll res = 0;
    ll ele = b;
    for(int i = 0; i < n; i++){
        ll pre = sum(ele+1);
        res += ele - pre;
        add(ele+1,1);
        ele = (ele+a)%n;
    }
    return res;
}

int main()
{
#ifdef local
    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
#endif // local
    int cas = 0;
    while(~scanf("%I64d%I64d%I64d",&n,&a,&b)&&n){
        memset(C,0,sizeof(C));
        printf("Case %d: %I64d\n",++cas,Count());
    }

    return 0;
}