组合数与优化算法-优快云博客

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;
}

组合数证明

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;
}

转载于:https://www.cnblogs.com/Stephen-F/p/9884660.html