“玲珑杯”ACM比赛 Round #11

A题：

思路：暴力一下就行，用long long，不然有可能会溢出，最好用C++，java有时候正确，有时候编译错误

AC代码：

#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;//2139062143typedef 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题：
思路：标称上是打表，我是用一个数组进行标记的，只要以前出现过这个数就退出，不然肯定会陷入死循环
AC代码：
#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;//2139062143typedef 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;    }}