3419. 假水仙花数

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我的想法比较朴素，就是一个个找。代码如下：

    using namespace std;
    inline unsigned four_times(unsigned i)
    {
        return i * i * i * i;
    }
    inline bool waterlilly(unsigned const& is)
    {
        unsigned i = is, temp = 0;
        while (i)
        {
            temp += four_times(i % 10);
            i /= 10;
        }
        if (is == temp)
            return true;
        return false;
    }


    int main()
    {
        vector<unsigned> vuzu;
        for (unsigned i = 0; i < 10000; ++i)
        {
            if (waterlilly(i))
                vuzu.push_back(i);
        }
        for (auto const& i : vuzu)
            cout << i << ' ';
    }

include

using namespace std;

bool isok(int n){
int sum = 0, s = n;
while(n > 0){
int left = n % 10;
sum += left * left * left * left;
n /= 10;
}
if(sum == s) return true;
return false;
}

int main(){
for(int i = 0; i < 10000; i ++)
if(isok(i))
cout << i << endl;
return 0;
}

include

include

int main()
{
int i,a,b,c,d;
printf(“0\n1\n”);
for(i=11;i<10000;i++)
{
if(i<100)
{
a=i%10;
b=i/10;
if(a<5&&b<5) continue;
else if(aaaa+bbbb==i) printf(“%d\n”,i);
}
if(i>100&&i<1000)
{
a=i%10;
b=i/10%10;
c=i/100;
if(a<5&&b<5&&c<5) continue;
else if(aaaa+bbbb+cccc==i) printf(“%d\n”,i);
}
if(i>=1000&&i<10000)
{
a=i%10;
b=i/10%10;
c=i/100%10;
d=i/1000;
if(a<5&&b<5&&c<5&&d<5) continue;
else if(aaaa+bbbb+cccc+ddd*d==i) printf(“%d\n”,i);
}

}
return 0;

}

include

int main()
{

for(int i=0;i<10000;i++)
{
    int sum=0;
    for(int j=i;j>0;j/=10)
    {
        sum+=(j%10)*(j%10)*(j%10)*(j%10);
    }
    if(sum==i)
        printf("%d\n",i);
}

}