完美立方：
//a的三次方等于a，b，c三个数的立方之和，这一组数称为完美立方
#includeusing namespace std;
int main()
{
int n;
cin >> n;
int count = 0;
for (int a = 2; a <= n; a++)
{
for (int b = 2; b < a; b++)
{
for (int c = b; c < a; c++)
{
for (int d = c; d < a; d++)
{
if (a*a*a == b*b*b + c*c*c + d*d*d)
{
printf("Cube=%d,Tripe=(%d,%d,%d)", a, b, c, d);
}
count++;
}
}
}
}
cout << "times:" << count;
}
求人体的生理高峰的日子：
#includeusing namespace std;
#define N 21252
int main()
{
//人体的体力智商情商高峰分别每23，28，33天出现一次，a，b，c分别是三次高峰出现的日子，求下一次高峰出现的日子
//int a, b, c;
//cin >> a >> b >> c;
//for (int i = 0; i < 21252; i++) //最简单的枚举法
//{
// if ((i - a) % 23 == 0 && (i - b) % 28 == 0 && (i - c) % 33 == 0)
// {
// cout << i;
// }
//}
int p, e, i, d, caseNo = 0;
while (cin>>p>>e>>i>>d&&p!=-1)
{
++caseNo;
int k;
for (k = d + 1; (k - p) % 23; ++k);
for (; (k - e) % 28; k += 23);
for (; (k - i) % 33; k += 23 * 28);
cout << "case" << caseNo << "下一次高峰出现在" << k - d << "天后";
}
return 0;
}
分析出假币并判断出假币的重量：
#includeusing namespace std;
//若干硬币，有一枚假币，称量三次，找出假币
char Left[3][7]; //天平左边硬币
char Right[3][7]; //天平右边硬币
char Result[3][7];
bool IsFake(char c, bool light) //light表示假币为轻，否则表示假币为重
{
for (int i = 0; i < 3; i++)
{
char *pLeft, *pRight;
if (light)
{
*pLeft = Left[0][0];
pRight = Right[i];
}
else {
pLeft = Left[i];
pRight = Right[i];
}
switch (Result[i][0])
{
case 'u':
if (strchr(pRight, c) == NULL)
{
return false;
}
break;
case 'e':
if (strchr(pLeft, c) || strchr(pRight, c))
{
return false;
}
break;
case 'd':
if (strchr(pLeft, c) == NULL)
{
return false;
}
break;
}
}
return true;
}
int main()
{
int t;
cin >> t;
while (t--)
{
for (int i = 0; i < 3; i++)
{
cin >> Left[i] >> Right[i] >> Result[i];
for (char c = 'A'; c <= 'L'; c++)
{
if (IsFake(c, true))
{
cout << c << "是假币并且比较轻" << endl;
}
else {
if (IsFake(c, false))
{
cout << c << "是假币并且比较重" << endl;
}
}
}
}
}
}
