HDU3030 Increasing Speed Limits 树状数组

http://acm.hdu.edu.cn/showproblem.php?pid=3030

该题题义就是给定一个长度为M的A数组，依照这个数组生成一个长度为N的数组，然后求严格增长的子串个数。

对于所给定的值进行离散化排序，再巧用树状数组求个数。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <map>
#include <algorithm>
#define MOD 1000000007
#define MAXN 500005
using namespace std; 

int num[MAXN], A[MAXN], c[MAXN], seq[MAXN], cnt;

inline int lowbit(int x)
{
    return x & -x;
}


inline void modify(int pos, int val)
{
    for (int i = pos; i <= cnt; i += lowbit(i))
    { 
        c[i] += val;
        if (c[i] >= MOD)
            c[i] %= MOD;
    }
}

inline int getsum(int pos)
{
    int s = 0;
    for (int i = pos; i > 0; i -= lowbit(i))
    {
        s += c[i];
        if (s >= MOD)
            s %= MOD;
    }
    return s;
}

int main()
{
    int T, n, m;
    long long X, Y, Z, res; 
    scanf("%d", &T);
    for (int ca = 1; ca <= T; ++ca)
    {
        res = 0;
        map<int,int>mp;
        scanf("%d %d %I64d %I64d %I64d", &n, &m, &X, &Y, &Z);
        memset(c, 0, sizeof (c));
        for (int i = 0; i < m; ++i)
        {
            scanf("%d", &A[i]);
        }
        for (int i = 0; i < n; ++i)
        {
            num[i+1] = A[i%m];
            seq[i] = num[i+1];
            A[i%m] = (X*A[i%m]+Y*(i+1))%Z;
        }
        sort(num+1, num+1+n);
        cnt = unique(num+1, num+1+n) - (num+1);
        for (int i = 1; i <= cnt; ++i)
        {
            mp[num[i]] = i;
        }
        modify(1, 1);
        for (int i = 0; i < n; ++i)
        { 
            long long x = getsum(mp[seq[i]]);   
            res += x;
            if (res > MOD)
                res %= MOD;
            modify(mp[seq[i]]+1, x);
        }
        printf("Case #%d: %I64d\n", ca, res);
    }
    return 0;
}