“玲瓏杯”ACM比賽 Round #11

A題：

思路：暴力一下就行，用long long，不然有可能會溢出，最好用C++，Java有時候正確，有時候編譯錯誤

AC代碼：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>

const int inf = 0x7f7f7f7f;//2139062143
typedef long long ll;
using namespace std;

int main()
{
    ll a,b,n;
    while (scanf("%I64d%I64d%I64d",&a,&b,&n) != EOF) {
        int index = 0;
        ll min1 = LONG_LONG_MAX;
        for (int i=1; i<=n; i++) {
            if (n % i == 0) {
                ll sum = n / i * a + b * n * i;
                if (sum < min1) {
                    min1 = sum;
                    index = i;
                }
            }
        }
        printf("%d\n",index);
    }
    return 0;
}

D題：

思路：標稱上是打表，我是用一個數組進行標記的，只要以前出現過這個數就退出，不然肯定會陷入死循環

AC代碼：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>

const int inf = 0x7f7f7f7f;//2139062143
typedef long long ll;
using namespace std;


int getSum(int n)
{
    int sum = 0;
    while (n != 0) {
        int temp = n % 10;
        sum += temp * temp;
        n /= 10;
    }
    return sum;
}
int main()
{
    int n;
    int a[10000];
    while(scanf("%d",&n) != EOF)
    {
        int sum = n;
        memset(a,0,sizeof(a));
        bool flag = false;
        while (true) {
            sum = getSum(sum);
            if (sum == 1) {
                flag = true;
                break;
            }
            if (a[sum] > 0)
                break;
            a[sum] = 1;
        }
        if (flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
}