传送门啦

15分暴力，但看题解说暴力分有30分。

就是找到公式，然后套公式。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;long long read(){char ch;bool f = false;while((ch = getchar()) < '0' || ch > '9')if(ch == '-')  f = true;int res = ch - 48;while((ch = getchar()) >= '0' && ch <='9')res = res * 10 - ch + 48;return f ? res + 1 : res;
}long long jc(long long a){//求阶乘 if(a == 0)  return 1;long long ans = 1;for(int i=1;i<=a;i++)ans *= i;return ans; //b = !a
} long long C(long long n,long long m){return jc(n) / (jc(m) * jc(n - m));
}
//组合数公式：Cn^m = !n / (!m * !(n - m)) long long t,k,n,m;
long long sum,x;int main(){t = read();  k = read();while(t--){x = 0;n = read();  m = read();//sum = jc(n) / (jc(m) * jc(n - m));for(long long i=1;i<=n;i++){//int j = min(i , m);for(long long j=1;j<=min(i,m);j++){//sum = jc(i) / (jc(j) * jc(i - j));if(C(i,j) % k == 0)x++;}}printf("%lld\n",x);}return 0;
}

15分，我现在用了组合数的递推公式，按理说应该更快了，但。。（想不通，数据范围在那里啊）

c[i][j]即为从i件物品中选j件的方案数。如果第i件物品不选，方案数就变为c[i-1][j]，如果选第i件物品，方案数就变为c[i-1][j-1]，总方案数就为两种情况的方案数之和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 2005;long long read(){char ch;bool f = false;while((ch = getchar()) < '0' || ch > '9')if(ch == '-')  f = true;int res = ch - 48;while((ch = getchar()) >= '0' && ch <='9')res = res * 10 - ch + 48;return f ? res + 1 : res;
}long long t,k,n,m;
long long sum,x,C[maxn][maxn];int main(){t = read();  k = read();while(t--){x = 0;C[1][0] = C[1][1] = 1;n = read();  m = read();for(long long i=2;i<=n;i++){C[i][0] = 1;for(long long j=1;j<=min(i,m);j++){C[i][j] = C[i-1][j] + C[i-1][j-1];if(C[i][j] % k == 0)x++;}}printf("%lld\n",x);}return 0;
}

为了提高效率，我们可以进行进一步的优化，就是预处理出组合数从而求出所有区间的满足条件的组合数个数，这里就要用到二维前缀和

杨辉三角

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 2005;inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}long long t,k,n,m;
long long sum[maxn][maxn],x,C[maxn][maxn];void work(){for(int i=1;i<=2000;i++){C[i][0] = 1;C[i][i] = 1;}C[1][1] = 1;for(long long i=2;i<=2000;i++)for(long long j=1;j<i;j++){C[i][j] = (C[i-1][j] + C[i-1][j-1]) % k;}for(long long i=1;i<=2000;i++){for(long long j=1;j<=i;j++){sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];if(C[i][j] == 0)sum[i][j]++ ;}sum[i][i+1] = sum[i][i]; }
}int main(){memset(C,0,sizeof(C));memset(sum,0,sizeof(sum));t = read();  k = read();work();while(t--){n = read();  m = read();m = min(n , m);printf("%lld\n",sum[n][m]);}return 0;
}

还有一个事不得不说，我改了一下午竟然发现是自己的快读打错了：

修改后：

暴力 40分

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}long long jc(long long a){//求阶乘 if(a == 0)  return 1;long long ans = 1;for(int i=1;i<=a;i++)ans *= i;return ans; //b = !a
} long long C(long long n,long long m){return jc(n) / (jc(m) * jc(n - m));
}
//组合数公式：Cn^m = !n / (!m * !(n - m)) long long t,k,n,m;
long long sum,x;int main(){t = read();  k = read();while(t--){x = 0;n = read();  m = read();//sum = jc(n) / (jc(m) * jc(n - m));for(long long i=1;i<=n;i++){//int j = min(i , m);for(long long j=1;j<=min(i,m);j++){//sum = jc(i) / (jc(j) * jc(i - j));if(C(i,j) % k == 0)x++;}}printf("%lld\n",x);}return 0;
}

递推公式 70

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 2005;inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}long long t,k,n,m;
long long sum,x,C[maxn][maxn];int main(){t = read();  k = read();while(t--){x = 0;C[1][0] = C[1][1] = 1;n = read();  m = read();for(long long i=2;i<=n;i++){C[i][0] = 1;for(long long j=1;j<=min(i,m);j++){C[i][j] = C[i-1][j] + C[i-1][j-1];if(C[i][j] % k == 0)x++;}}printf("%lld\n",x);}return 0;
}