24点 C语言程序

在网上找了个代肢隐码，我改了下，符合你历友厅的三个要求了。

#include<stdio.h>

double

fun(double

a1,double

a2,int

b)

{switch(b)

{case

0:return

(a1+a2)

case

1:return

(a1-a2)

case

2:return

(a1*a2)

case

3:return

(a1/a2)

}

}

void

main()

{int

i,j,k,l,n,m,r,save[4],flg=1

double

num[4]={1,1,1,1},tem1,tem2,tem3,abc=1111

char

sign[5]="+-*/"

printf("输入四个告液数:")

for(i=0i<4i++)

{scanf("%lf",num+i)

save[i]=num[i]if(save[i]>13)flg=0}

if(flg)

{

flg=0

for(i=0i<4i++)

for(j=0j<4j++)

if(j!=i)

{for(k=0k<4k++)

if(k!=i&&k!=j)

{for(l=0l<4l++)

if(l!=i&&l!=j&&l!=k)

{for(n=0n<4n++)

for(m=0m<4m++)

for(r=0r<4r++)

{tem1=fun(num[i],num[j],n)

tem2=fun(tem1,num[k],m)

tem3=fun(tem2,num[l],r)

if(tem3==24.0)

{printf("{(%d%c%d)%c%d}%c%d=24\n",save[i],sign[n],save[j],sign[m],save[k],sign[r],save[l])return}

else

if(tem3==-24.0)

{printf("{%d%c(%d%c%d)}%c%d=24\n",save[k],sign[m],save[i],sign[n],save[j],sign[r],save[l])return}

else

if(tem3==1.0/24.0)

{printf("%d%c{(%d%c%d)%c%d}=24\n",save[l],sign[r],save[i],sign[n],save[j],sign[m],save[k])return}

else

if(tem3==-1.0/24.0)

{printf("%d%c{%d%c(%d%c%d)}=24\n",save[l],sign[r],save[k],sign[n],save[i],sign[m],save[j])return}

else

{tem1=fun(num[i],num[j],n)

tem2=fun(num[k],num[l],r)

tem3=fun(tem1,tem2,m)

if(tem3==24.0)

{printf("(%d%c%d)%c(%d%c%d)=24\n",save[i],sign[n],save[j],sign[m],save[k],sign[r],save[l])return}

}

}

}

}

}

}

if(!flg)

printf("NO

ANSWER\n")

}

#include<iostream>

using namespace std

struct operation

{

int x

int y

int o

}op[3]

bool co=false

void g(int a[],int b[],int n,int i,int j)

{

int x,y

for(x=0,y=1x<nx++)

{

if(x==i||x==j)continue

b[y++]=a[x]

}

op[4-n].x=a[i]

op[4-n].y=a[j]

}

void f(int a[],int n)

{

int i,j,t,b[4]

if(n==1&&a[0]==24)co=true

for(i=0i<n-1i++)

{

for(j=i+1j<nj++)

{

for(t=1t<=6t++)

{

switch(t)

{

case 1:g(a,b,n,i,j)

b[0]=a[i]+a[j]

op[4-n].o=t

f(b,n-1)

if(co)return

break

case 2:g(a,b,n,i,j)

b[0]=a[i]-a[j]

op[4-n].o=t

f(b,n-1)

if(co)return

break

case 3:g(a,b,n,i,j)

b[0]=a[i]*a[j]

op[4-n].o=t

f(b,n-1)

if(co)return

break

case 4:

if(a[j]!=0&&a[i]%a[j]==0)

{

g(a,b,n,i,j)

b[0]=a[i]/a[j]

op[4-n].o=t

f(b,n-1)

if(co)return

}

break

case 5:g(a,b,n,i,j)

b[0]=a[j]-a[i]

op[4-n].o=t

f(b,n-1)

if(co)return

break

case 6:

if(a[i]!=0&&a[j]%a[i]==0&&a[j]>a[i])

{

g(a,b,n,i,j)

b[0]=a[j]/a[i]

op[4-n].o=t

f(b,n-1)

if(co)return

}

break

}

}

}

}

}

void h(int a[],int n,int x,int y,int h)

{

int i,j,b[4]

bool bx=false,by=false

for(i=0,j=0i<ni++)

{

if(a[i]==x&&bx==false)

{

bx=true

continue

}

if(a[i]==y&&by==false)

{

by=true

continue

}

b[j++]=a[i]

}

for(i=0i<n-1i++)a[i+1]=b[i]

a[0]=h

}

void main()

{

int a[4],i,j

cout<<"输入4个整数:"

for(i=0i<4i++)cin>>a[i]

f(a,4)

if(co)

{

for(i=0i<3i++)

{

for(j=0j<4-ij++)cout<<a[j]<袜闭<" "

cout<<"\n运算 "

switch(op[i].o)

{

case 1:cout<<op[i].x<<"+"仔滚<<op[i].y<<endl

h(a,4-i,op[i].x,op[i].y,op[i].x+op[i].y)

break

case 2:cout<<op[i].x<<"告戚裂-"<<op[i].y<<endl

h(a,4-i,op[i].x,op[i].y,op[i].x-op[i].y)

break

case 3:cout<<op[i].x<<"*"<<op[i].y<<endl

h(a,4-i,op[i].x,op[i].y,op[i].x*op[i].y)

break

case 4:cout<<op[i].x<<"/"<<op[i].y<<endl

h(a,4-i,op[i].x,op[i].y,op[i].x/op[i].y)

break

case 5:cout<<op[i].y<<"-"<<op[i].x<<endl

h(a,4-i,op[i].x,op[i].y,op[i].y-op[i].x)

break

case 6:cout<<op[i].y<<"/"<<op[i].x<<endl

h(a,4-i,op[i].x,op[i].y,op[i].y/op[i].x)

break

}

}

cout<<"24"<<endl

}

else cout<<"不能算出24"<<endl

}

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