c++实现计算二十四点–zj
首先，要感谢vaintwyt的C++实现24点算法，使我有了灵感，编出了计算24点的程序。
不懂的，直接去刚才第二个链接里看。
上代码
#include<bits/stdc++.h>
#define f(x) for(x=1;x<=4;x++)
using namespace std;
map<string,int> q;
char o[5]="+-*/";
int tot=0;
bool flag=false;
double myabs(double x){return x>0?x:-x;}
bool check(double x){return myabs(x-24)<0.000001;}
double Add(double a, double b){return a+b;}
double Sub(double a, double b){return a-b;}
double Mul(double a, double b){return a*b;}
double Div(double a, double b){if(!b)return -100000;return a/b;}
double(*op[4])(double,double) = {Add, Sub, Mul, Div};//函数指针。C++确实强大
void Game24Points(double a, double b, double c, double d)
{
for(int i=0; i<4; i++)
{
for(int j=0; j<4; j++)
{
for(int k=0; k<4; k++)
{
//((a#b)#c)#d
double ret = op[k](op[j](op[i](a, b), c) , d);
if(check(ret))
{
printf("\n第%d种:\n((%0.0lf%c%0.0lf)%c%0.0lf)%c%0.0lf=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//(a#(b#c))#d
ret = op[k](op[i](a, op[j](b, c)) , d);
if(check(ret))
{
printf("\n第%d种:\n(%0.0lf%c(%0.0lf%c%0.0lf))%c%0.0lf=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//a#((b#c)#d)
ret = op[i](a, op[k](op[j](b, c) , d));
if(check(ret))
{
printf("\n第%d种:\n%0.0lf%c((%0.0lf%c%0.0lf)%c%0.0lf)=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//(a#b)#(c#d)
ret = op[j](op[i](a, b), op[k](c, d));
if(check(ret))
{
printf("\n第%d种:\n(%0.0lf%c%0.0lf)%c(%0.0lf%c%0.0lf)=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//a#(b#(c#d))
ret = op[i](a, op[j](b, op[k](c, d)));
if(check(ret))
{
printf("\n第%d种:\n%0.0lf%c(%0.0lf%c(%0.0lf%c%0.0lf))=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
}
}
}
}
int main()
{
double a[5];
int i,j,k;
f(i)cin>>a[i];
bool s=false;
f(i)
{
f(j)
{
if(i==j)continue;
f(k)
{
if(i==k||i==10-i-j-k||j==k||j==10-j-k-i||k==10-j-i-k)
continue;
string a1,b1,c1,d1,f1;
a1=char(a[i]+'0');b1=char(a[j]+'0');c1=char(a[k]+'0');d1=char(a[10-i-j-k]+'0');
f1=a1+' '+b1+' '+c1+' '+d1;
if(q[f1])
continue;
q[f1]=1;
Game24Points(a[i],a[j],a[k],a[10-i-j-k]);
}
}
}
if(flag)
printf("\n共%d个解\n",tot);
else printf("此题无解\n");
return 0;
}
接着，我又想。
1 2 2 3不是可以**(2^(2+1))*3**的吗？我就添加了乘方和对数运算。
代码
#include<bits/stdc++.h>
#define f(x) for(x=1;x<=4;x++)
using namespace std;
map<string,int> q;
char o[8]="+-*/^lg";
int tot=0;
bool flag=false;
double myabs(double x){return x>0?x:-x;}
double mylog(double a,double b)
{
if(b==1)return 0;
if(a<=1&&a<b)return -100000;
int ans=0;
double t=1;
while(t<b)
{
t*=a;
ans++;
}
if(t==b)return ans;
return -100000;
}
bool check(double x){return myabs(x-24)<0.000001;}
double Add(double a, double b){return a+b;}
double Sub(double a, double b){return a-b;}
double Mul(double a, double b){return a*b;}
double Div(double a, double b){if(!b)return -100000;return a/b;}
double Pow(double a, double b){return pow(a,b);}
double Log(double a, double b){return mylog(a,b);}
double Sqrt(double a,double b)
{
int l=0,r=b+1,mid;
while(l+1<r)
{
mid=(l+r)>>1;
if(pow(mid,a)>=b)r=mid;
else l=mid;
}
if(pow(r,a)==b)return r;
return -100000;
}
double(*op[7])(double,double) = {Add, Sub, Mul, Div, Pow, Log, Sqrt};//函数指针。C++确实强大
void Game24Points(double a, double b, double c, double d)
{
for(int i=0; i<7; i++)
{
for(int j=0; j<7; j++)
{
for(int k=0; k<7; k++)
{
//((a#b)#c)#d
double ret = op[k](op[j](op[i](a, b), c) , d);
if(check(ret))
{
printf("\n第%d种:\n((%0.0lf%c%0.0lf)%c%0.0lf)%c%0.0lf=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//(a#(b#c))#d
ret = op[k](op[i](a, op[j](b, c)) , d);
if(check(ret))
{
printf("\n第%d种:\n(%0.0lf%c(%0.0lf%c%0.0lf))%c%0.0lf=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//a#((b#c)#d)
ret = op[i](a, op[k](op[j](b, c) , d));
if(check(ret))
{
printf("\n第%d种:\n%0.0lf%c((%0.0lf%c%0.0lf)%c%0.0lf)=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//(a#b)#(c#d)
ret = op[j](op[i](a, b), op[k](c, d));
if(check(ret))
{
printf("\n第%d种:\n(%0.0lf%c%0.0lf)%c(%0.0lf%c%0.0lf)=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
//a#(b#(c#d))
ret = op[i](a, op[j](b, op[k](c, d)));
if(check(ret))
{
printf("\n第%d种:\n%0.0lf%c(%0.0lf%c(%0.0lf%c%0.0lf))=24\n",++tot,a,o[i],b,o[j],c,o[k],d);
flag=true;
}
}
}
}
}
int main()
{
printf("请输入你要进行运算的4个数:\n");
double a[5];
int i,j,k;
f(i)printf("\n第%d个:",i),cin>>a[i];
bool s=false;
f(i)
{
f(j)
{
if(i==j)continue;
f(k)
{
if(i==k||i==10-i-j-k||j==k||j==10-j-k-i||k==10-j-i-k)
continue;
string a1,b1,c1,d1,f1;
a1=char(a[i]+'0');b1=char(a[j]+'0');c1=char(a[k]+'0');d1=char(a[10-i-j-k]+'0');
f1=a1+' '+b1+' '+c1+' '+d1;
if(q[f1])
continue;
q[f1]=1;
Game24Points(a[i],a[j],a[k],a[10-i-j-k]);
}
}
}
if(flag)
{
printf("\n共%d个解\n",tot);
printf("\n温馨提示:\n1.\"^\"为乘方运算,a^b=a的b次方\n2.\"l\"为对数运算,alb=log a(b)\n3.\"g\"为开方运算,agb=b的a次方根\n");
}
else printf("此题无解\n");
return 0;
}
谢谢–zj
