題目大意:
已知
思路:
可以枚舉
如果爲沒有出現過的數,即該數與枚舉到當前情況的積互素,乘以
否則乘以
即:
#include <iostream>
#include <cstring>
#include <cstdio>
#define LL long long
#define N 502
#define M 4000
#define MOD 1000000007
using namespace std;
bool not_prime[M];
int prim[N*2];
LL dp[N][N];
void init()
{
memset(not_prime, 0, sizeof(not_prime));
memset(dp, 0, sizeof(dp));
prim[1] = 0;
int cnt = 0;
for(int i = 2; i < M; i++)
{
if(!not_prime[i])
{
prim[++cnt] = i;
for(int j = i + i; j < M; j += i)
{
not_prime[j] = 1;
}
}
}
dp[0][0] = 1;
for (int i = 1; i <= 500; i++)
{
for (int j = 1; j <= i; j++)
{
dp[i][j] = (dp[i][j] + dp[i-1][j] * prim[j]) % MOD;
dp[i][j] = (dp[i][j] + dp[i-1][j-1] * (prim[j] - 1)) % MOD;
}
}
}
int main()
{
init();
int T;
scanf("%d", &T);
for (int cas = 1; cas <= T; cas++)
{
int n, m;
scanf("%d%d", &n, &m);
printf("Case %d: %lld\n", cas, dp[n][m]);
}
return 0;
}