1 条题解

  • 0
    @ 2021-6-14 23:25:23

    C :

    #include<stdio.h>
    void main()
    {
        int a, n, m, x;
        scanf("%d%d%d%d",&a,&n,&m,&x);
        int f[24];
        f[1] = 1;
        f[2] = 1;
        for(int i = 3; i < 24; i++)
        {
            f[i] = f[i-1] + f[i-2];
        }
        printf("%d\n",(f[x-2]+1)*a+(f[x-1]-1)*(m-(f[n-3]+1)*a)/(f[n-2]-1));
    } 
    
    

    C++ :

    #include<stdio.h>
    int main()
    {
    	int a,n,m,x,k,i,b[100],c[100],d[100];
    	while(scanf("%d%d%d%d",&a,&n,&m,&x)!=EOF)
    	{
    		if(n<=3)
    		{
    			printf("%d\n",a);
    			continue;
    		}
    		if(x==n)
    		{
    			printf("0\n");
    			continue;
    		}
    		k=1;
    		b[1]=a,b[2]=a;
    		c[1]=a,d[1]=0;
    		b[n-1]=0;
    		while(b[n-1]!=m)
    		{
    		
    			c[2]=k,d[2]=k;
    			for(i=3;i<=n-1;i++)
    			{
    				c[i]=c[i-1]+c[i-2];
    				d[i]=c[i-1];
    				b[i]=b[i-1]+c[i]-d[i];
    			}
    			k++;
    		}
    		printf("%d\n",b[x]);
    	}
    	return 0;
    }
    

    Pascal :

    var
     a,n,x,m,s,i:longint;
     a1:array[1..2,1..20] of longint;
     a2:array[1..2,1..20] of longint;
     a3:array[1..2,1..20] of int64;
    begin
     readln(a,n,m,x);
     a1[1,1]:=1;a1[1,2]:=0; a1[2,1]:=0;a1[2,2]:=1;
     a2[1,1]:=0;a2[1,2]:=0; a2[2,1]:=0;a2[2,2]:=1;
     a3[1,1]:=1;a3[1,2]:=1;
      for i:=3 to 20 do
       begin
       a1[1,i]:=a1[1,i-1]+a1[1,i-2];
       a1[2,i]:=a1[2,i-1]+a1[2,i-2];
       a2[1,i]:=a1[1,i-1];
       a2[2,i]:=a1[2,i-1];
       a3[1,i]:=a3[1,i-1]+a1[1,i]-a2[1,i];
       a3[2,i]:=a3[2,i-1]+a1[2,i]-a2[2,i];
       end;
      s:=(m-a3[1,n-1]*a) div a3[2,n-1];
      writeln(a3[1,x]*a+a3[2,x]*s);
    end.
    
    

    Python :

    # coding=utf-8
    #a,n,m,x=5,7,32,4
    a,n,m,x=map(int,input().split())
    ls1=[[1,0],[1,0],[2,0]]
    ls2=[[1,0],[0,1],[1,1]]
    for i in range(3,n):
    	an2=ls2[i-1][0]+ls2[i-2][0]
    	bn2=ls2[i-1][1]+ls2[i-2][1]
    	ls2.append([an2,bn2])
    	an1=ls1[i-1][0]+ls2[i-2][0]
    	bn1=ls1[i-1][1]+ls2[i-2][1]
    	ls1.append([an1,bn1])
    b=(m-ls1[n-2][0]*a)//ls1[n-2][1]
    print(ls1[x-1][0]*a+ls1[x-1][1]*b)
    
    • 1

    信息

    ID
    190
    时间
    1000ms
    内存
    125MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者