第1题
#include <iostream>
using namespace std;
int main()
{
freopen("apple.in", "r", stdin);
freopen("apple.out", "w", stdout);
int a, b;
cin >> a >> b;
cout << (a % b ? (a / b + 1) : a / b) << endl;
return 0;
}
第2题
#include <iostream>
using namespace std;
int main()
{
// freopen("height.in", "r", stdin);
// freopen("height.out", "w", stdout);
int n;
cin >> n;
int mx = 0, secMx = 0;
int mxPos = 0, secPos = 0;
int a[n + 1];
for(int i = 1; i <= n; i++)
{
cin >> a[i];
if(a[i] > mx)
{
secMx = mx;
secPos = mxPos;
mx = a[i];
mxPos = i;
}
else if(a[i] > secMx)
{
secMx = a[i];
secPos = i;
}
}
cout << secPos << ' ' << secMx << endl;
return 0;
}
第3题
#include <iostream>
using namespace std;
int main()
{
// freopen("change.in", "r", stdin);
// freopen("change.out", "w", stdout);
int n;
cin >> n;
int change = 100 - n;
int cnt = 0;
cnt += change / 50;
change %= 50;
cnt += change / 20;
change %= 20;
cnt += change / 10;
change %= 10;
cnt += change / 5;
change %= 5;
// 最后加上一元钱的数量
cnt += change;
cout << cnt << endl;
return 0;
}
扩展：假如有三种面值纸币：1元，5元，11元（现实中没有11元的币值，但是不要纠结，只是做个假设而已）。找零15元，至少需要几张纸币？
完整答案请加微信307591841或QQ307591841
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