HDU 5565 Clarke and baton

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke split into $n$ guys, named $1$ to $n$.
They will play a game called Lose Weight. Each of Clarkes has a weight $a[i]$. They have a baton which is always in the hand of who has the largest weight(if there are more than 2 guys have the same weight, choose the one who has the smaller order). The one who holds the baton needs to lose weight. i.e. $a[i]$ decreased by 1, where$i$ is the guy who holds the baton.
Now, Clarkes know the baton will be passed $q$ times. They want to know each one's weight after finishing this game.

Input

The first line contains an integer $T(1\le T\le 10)$, the number of the test cases.
Each test case contains three integers n, q, seed(1 \le n, q \le 10000000, \sum n \le 20000000, 1 \le seed \le 10^9+6)n,q,seed(1≤n,q≤10000000,∑n≤20000000,1≤seed≤10​9​​+6), $seed$ denotes the random seed.
$a[i]$ generated by the following code.

long long seed;
int rand(int l, int r) {
    static long long mo=1e9+7, g=78125;
    return l+((seed*=g)%=mo)%(r-l+1);
}

int sum=rand(q, 10000000);
for(int i=1; i<=n; i++) {
    a[i]=rand(0, sum/(n-i+1));
    sum-=a[i];
}
a[rand(1, n)]+=sum;

Output

Each test case print a line with a number which is $(b[1]+1)xor(b[2]+2)xor...xor(b[n]+n)$, where $b[i]$ is the $ith$ Clarke's weight after finishing the game.

Sample Input

1
3 2 1

Sample Output

20303

Hint:
At first, $a[1]=20701,a[2]=31075,a[3]=26351$.  
At the first time, $a[2]$ decrease by 1.  
At the second time, $a[2]$ decrease by 1.  
Finally, $a[1]=20701,a[2]=31073,a[3]=26351$. So answer is (20701+1)xor(31073+2)xor(26351+3)=20303
$(20701+1)xor(31073+2)xor(26351+3)=20303$. 

模拟一下操作就好了
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 10000005;
const int base = 1e9 + 7;
int T, limit;
int n, q, a[maxn], ft[maxn], nt[maxn];
LL seed;

int rand(int l, int r) {
    static long long mo = 1e9 + 7, g = 78125;
    return l + ((seed *= g) %= mo) % (r - l + 1);
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d%I64d", &n, &q, &seed);
        int sum = rand(q, 10000000);
        limit = 0;
        for (int i = 1; i <= n; i++)
        {
            a[i] = rand(0, sum / (n - i + 1));
            sum -= a[i];
            limit = max(a[i], limit);
        }
        limit = max(limit, a[rand(1, n)] += sum);
        for (int i = 1; i <= max(limit, n); i++) ft[i] = nt[i] = 0;
        for (int i = 1; i <= n; i++)
        {
            if (ft[a[i]])
            {
                nt[i] = ft[a[i]];
                ft[a[i]] = i;
            }
            else ft[a[i]] = i;
        }
        int tot = 0;
        for (int i = limit, j; i; i--)
        {
            if (q >= tot) { q = q - tot; limit = i; }
            else break;
            for (j = ft[i]; j; j = nt[j]) tot++;
        }
        int ans = 0;
        for (int i = 1; i <= n; i++)
        if (a[i] >= limit) if (q) { ans ^= (limit + i - 1); q--; }
        else ans ^= (limit + i);
        else ans ^= a[i] + i;
        printf("%d\n", ans);
    }
    return 0;
}