C 解法, 执行用时: 1ms, 内存消耗: 348KB, 提交时间: 2020-07-07

#include <stdio.h>
#include <stdlib.h>

int func(int x,int y)
{
	if(x==0||y==0)
		return 1;
	else
		return func(x-1,y)+func(x,y-1);
}

int main(void) {
	int x,y;
	scanf("%d %d",&x,&y);
	printf("%d",func(x,y));
}

C 解法, 执行用时: 2ms, 内存消耗: 280KB, 提交时间: 2021-08-09

// dfs + memorization

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


int depth_first_search_with_memorization_algorithm(int* dp,
                                                    const int m,
                                                    const int n,
                                                    int x,
                                                    int y) {
  if (x == n || y == m)         return 0;
  if (x == n - 1 && y == m - 1) return 1;
  if (dp[y * n + x] >= 0)       return dp[y * n + x];
  
  int count = 0;
  count += depth_first_search_with_memorization_algorithm(dp, m, n, x + 1, y); // right 
  count += depth_first_search_with_memorization_algorithm(dp, m, n, x, y + 1); // down
  return dp[y * n + x] = count;
}


int main(const int argc, const char** const argv) {
  int m, n;
  fscanf(stdin, "%d %d", &m, &n);
  ++m, ++n;
  
  int dp[m][n];
  memset(dp, -1, sizeof dp);
  
  fprintf(stdout, "%d\n",
          depth_first_search_with_memorization_algorithm(dp, m, n, 0, 0));
  return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 352KB, 提交时间: 2019-08-04

#include<stdio.h>

long long jiecheng(int n);
int C_fun(int a,int b);

int main()
{
	int x,y;
	scanf("%d%d",&x,&y);
	printf("%d",C_fun(x+y,x));
	// printf("20:%lld,10:%d\n",jiecheng(20),jiecheng(10));
	return 0;
}

long long jiecheng(int n)
{
	if(n==0||n==1)
		return 1;
	else
	{
		return n*jiecheng(n-1);
	}
	
}

int C_fun(int a,int b)
{
	return jiecheng(a)/jiecheng(a-b)/jiecheng(b);
}

C 解法, 执行用时: 2ms, 内存消耗: 352KB, 提交时间: 2019-07-03

#include<stdio.h>

long long a[] = {0,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600,6227020800,87178291200,1307674368000,20922789888000,355687428096000,6402373705728000,121645100408832000,2432902008176640000};

int main()
{
    int x,y;
    scanf("%d %d", &x, &y);
    printf("%lld", a[x+y]/(a[x]*a[y]));
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2020-05-20

#include<stdio.h>
 
long long jiecheng(int n);
int C_fun(int a,int b);
 
int main()
{
    int x,y;
    scanf("%d%d",&x,&y);
    printf("%d",C_fun(x+y,x));
    // printf("20:%lld,10:%d\n",jiecheng(20),jiecheng(10));
    return 0;
}
 
long long jiecheng(int n)
{
    if(n==0||n==1)
        return 1;
    else
    {
        return n*jiecheng(n-1);
    }
     
}
 
int C_fun(int a,int b)
{
    return jiecheng(a)/jiecheng(a-b)/jiecheng(b);
}