感悟：3道编程题，DFS+2道数论，第一次做类似leetcode的题目，有点不适应
现有一个 3x3 规格的 Android 智能手机锁屏程序和两个正整数 m 和 n ，请计算出使用最少m 个键和最多 n个键可以解锁该屏幕的所有有效模式总数。
其中有效模式是指：
1、每个模式必须连接至少m个键和最多n个键；
2、所有的键都必须是不同的；
3、如果在模式中连接两个连续键的行通过任何其他键，则其他键必须在模式中选择，不允许跳过非选择键（如图）；
4、顺序相关，单键有效（这里可能跟部分手机不同）。
输入：m,n
代表允许解锁的最少m个键和最多n个键
输出：
输出：
满足m和n个键数的所有有效模式的总数
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现给定任意正整数 n，请寻找并输出最小的正整数 m（m>9），使得 m 的各位（个位、十位、百位 ... ...）之乘积等于n，若不存在则输出 -1。
题解：迭代可以做，俺没想出来，硬算的
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在vivo产线上，每位职工随着对手机加工流程认识的熟悉和经验的增加，日产量也会不断攀升。
假设第一天量产1台，接下来2天(即第二、三天)每天量产2件，接下来3天(即第四、五、六天)每天量产3件 ... ...
以此类推，请编程计算出第n天总共可以量产的手机数量。
题解：手机量数就是 ： i ^ 2，所以只需要找到所在天数即可，一种是
for(i = 0; sum < n; i++) sum ++ i; i--;
还一种就是i *(i + 1) / 2 <= n，找到后求i ^ 2加起来即可
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JgGgBQDOSJss77r :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs .:J: :; .,:K6BQS7:.,.,. :r7Ji;r7;::;;. . **/ #include <bits/stdc++.h> using namespace std; #define FFI(a, b) for(int i = a; i < b; i++) #define FFJ(a, b) for(int j = a; j < b; j++) #define FFK(a, b) for(int k = a; k < b; k++) #define RR(a, b) for(int i = a; i > b; i--) #define ME(a, b) memset(a, b, sizeof(a)) #define FILL(a, b, c) fill(a, a + b, c) #define SC(x) scanf("%d", &x) #define SCC(a, b) scanf("%d%d", &a, &b) #define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c) #define PY() printf("Yes ") #define PN() printf("No ") #define PR(x) printf("%d ", x) #define PRR(a, b) printf("%d%d ", a, b); #define INF 0x3f3f3f3f #define MAX 2001 #define MOD 1000000007 #define E 2.71828182845 #define M 8 #define N 6 typedef long long LL; const double PI = acos(-1.0); typedef pair<int, int> Author; vector<pair<string, int> > VP; /*******************************************************************************************************************************************************/ int main(void){ ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int n, num[MAX], cnt, res, ans; ME(num, 0);ans = 0; for(int i = 1; i < MAX; i++){ num[i] += num[i - 1] + i * i; } cin>>n; for(int i = n - 1; i > 1; i--){ if(i * (i + 1) / 2 <= n){ cnt = i; res = n - i * (i + 1) / 2; break; } } ans += num[cnt] + (cnt + 1) * res; cout<<ans<<endl; return EXIT_SUCCESS; }