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(https://www.luogu.com.cn/problem/P2822)
30pts
直接上 long long, 暴力套公式…
代码就不给出了……
90pts
k∣C(i,j)⇔C(i,j) mod k=0
知道怎么做了吧
直接递推组合数并对 k 取模，判断这是不是0
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| #include <iostream>
#include <algorithm>
using namespace std;
const int N = 2e3 + 10;
int t, k;
int n, m;
int dp[N][N];
void build()
{
dp[0][0] = 1;
dp[1][0] = dp[1][1] = 1;
for(int i = 1; i <= 2000; i++)
{
dp[i][0] = 1;
for(int j = 1; j <= i; j++)
dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % k;
}
}
int main(void)
{
cin >> t >> k;
build();
for (int cse = 1; cse <= t; cse++)
{
scanf("%d%d", &n, &m);
int ans = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= min(i, m); j++)
{
if (!dp[i][j])
ans++;
}
printf("%d\n", ans);
}
}
|
100分
很多人看到这里已经知道了。。
没错，二维前缀和！
这里提供一种计算的新方法（看代码）:
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| #include <iostream>
#include <algorithm>
using namespace std;
const int N = 2e3 + 10;
int t, k;
int n, m;
int dp[N][N];
int s[N][N];
void build()
{
dp[0][0] = 1;
dp[1][0] = dp[1][1] = 1;
for(int i = 1; i <= 2000; i++)
{
dp[i][0] = 1;
for(int j = 1; j <= i; j++)
{
dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % k;
if(!dp[i][j]) s[i][j] = 1;
}
}
for(int i = 1; i <= 2000; i++) for(int j = 1; j <= 2000; j++) s[i][j] += s[i][j - 1];
for(int i = 1; i <= 2000; i++) for(int j = 1; j <= 2000; j++) s[i][j] += s[i - 1][j];
}
int main(void)
{
cin >> t >> k;
build();
for (int cse = 1; cse <= t; cse++)
{
scanf("%d%d", &n, &m);
printf("%d\n", s[n][m]);
}
}
|