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OpenJudge NOI 2.1 1749:数字方格
2022-07-06 07:12:00 【君义_noip】
【题目链接】
【题目考点】
1. 枚举
【解题思路】
用枚举方法求解方程组
- 枚举对象: a 1 , a 2 , a 3 a_1,a_2,a_3 a1,a2,a3
- 枚举范围: 0 ≤ a 1 , a 2 , a 3 ≤ n 0\le a_1, a_2, a_3 \le n 0≤a1,a2,a3≤n
- 枚举条件:
( a 1 + a 2 ) % 2 = 0 (a_1+a_2)\%2 = 0 (a1+a2)%2=0
( a 2 + a 3 ) % 3 = 0 (a_2+a_3)\%3 = 0 (a2+a3)%3=0
( a 1 + a 2 + a 3 ) % 5 = 0 (a_1+a_2+a_3)\%5 = 0 (a1+a2+a3)%5=0
枚举出来符合条件的 a 1 , a 2 , a 3 a_1,a_2,a_3 a1,a2,a3,选择 a 1 + a 2 + a 3 a_1+a_2+a_3 a1+a2+a3最大的那一组。
【题解代码】
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n, mx = 0;//mx:a1+a2+a3的最大值
cin >> n;
for(int a1 = 0; a1 <= n; ++a1)
for(int a2 = 0; a2 <= n; ++a2)
for(int a3 = 0; a3 <= n; ++a3)
{
if((a1+a2)%2 == 0 && (a2+a3)%3 == 0 && (a1+a2+a3)%5 == 0)
mx = max(mx, a1+a2+a3);
}
cout << mx;
return 0;
}
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