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所以我现在要通过项目欧拉我对问题17问题是项目欧拉任务17
如果数字1至5字写得出来:一,二,三,四,五,那么总共有3 + 3 + 5 + 4 + 4 = 19个字母。
如果所有从1到1000(单万)的数字包容性话写出来,多少个字母,会使用吗?
我个人,虽然这个问题将是非常直截了当。现在,我知道我的代码不是最优化的,我知道我可以在网络上找到解决方案。但我只是在寻找我的代码出了什么问题。我得到的答案是21224,而正确答案是21124.
int Sum = 0;
int Hundred;
int Ten;
int One;
void Count()
{
for(int i = 0; i < 10; i++)
{
Hundred = i;
for(int j = 0; j < 10; j++)
{
Ten = j;
for(int k = 0; k < 10; k++)
{
One = k;
switch(i)
{
case 0:
Tens();
break;
case 1:
Sum += 3;
Sum += 7;
Tens();
break;
case 2:
Sum += 3;
Sum += 7;
Tens();
break;
case 3:
Sum += 5;
Sum += 7;
Tens();
break;
case 4:
Sum += 4;
Sum += 7;
Tens();
break;
case 5:
Sum += 4;
Sum += 7;
Tens();
break;
case 6:
Sum += 3;
Sum += 7;
Tens();
break;
case 7:
Sum += 5;
Sum += 7;
Tens();
break;
case 8:
Sum += 5;
Sum += 7;
Tens();
break;
case 9:
Sum += 4;
Sum += 7;
Tens();
break;
}
}
}
}
Sum += 11 + 99*9*3; //Taking into account the "one thousand" and the "and"s
label1.Text = Sum.ToString();
}
void Tens()
{
switch (Ten)
{
case 0:
Ones();
break;
case 1:
switch (One)
{
case 0:
Sum += 3;
break;
case 1:
Sum += 6;
break;
case 2:
Sum += 6;
break;
case 3:
Sum += 8;
break;
case 4:
Sum += 8;
break;
case 5:
Sum += 7;
break;
case 6:
Sum += 7;
break;
case 7:
Sum += 9;
break;
case 8:
Sum += 8;
break;
case 9:
Sum += 8;
break;
}
break;
case 2:
Sum += 6;
Ones();
break;
case 3:
Sum += 6;
Ones();
break;
case 4:
Sum += 6;
Ones();
break;
case 5:
Sum += 5;
Ones();
break;
case 6:
Sum += 5;
Ones();
break;
case 7:
Sum += 7;
Ones();
break;
case 8:
Sum += 6;
Ones();
break;
case 9:
Sum += 6;
Ones();
break;
}
}
void Ones()
{
switch(One)
{
case 0:
break;
case 1:
Sum += 3;
break;
case 2:
Sum += 3;
break;
case 3:
Sum += 5;
break;
case 4:
Sum += 4;
break;
case 5:
Sum += 4;
break;
case 6:
Sum += 3;
break;
case 7:
Sum += 5;
break;
case 8:
Sum += 5;
break;
case 9:
Sum += 4;
break;
}
}