Codeforces Round #104 (Div. 2) A~D

博客内容涉及幸运数字的定义及判断,以及与票号相关的数学问题。包括判断一个数是否为幸运数字,找到最小的大于给定数且具有特定幸运数字特征的数,以及通过最少操作使两个幸运数字字符串相等的问题。
A. Lucky Ticket
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 477444 are lucky and 517467 are not.

Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals n (n is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first n / 2 digits) equals the sum of digits in the second half (the sum of the last n / 2 digits). Check if the given ticket is lucky.

Input

The first line contains an even integer n (2 ≤ n ≤ 50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly n — the ticket number. The number may contain leading zeros.

Output

On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).

Examples
input
Copy
2
47
output
NO
input
Copy
4
4738
output
NO
input
Copy
4
4774
output
YES
Note

In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).

### Codeforces Round 1005 Div. 2 A-F Problem Solutions #### A. Money Change 为了处理货币转换问题,可以将所有的金额都转化为分的形式来简化计算。通过遍历输入数据并累加各个部分的金额,最后求得剩余的钱数并对100取模得到最终结果[^2]。 ```cpp #include <iostream> using namespace std; int main() { int s, xi, yi; cin >> s; int total_cents = 0; for (int i = 0; i < s; ++i) { cin >> xi >> yi; total_cents += xi * 100 + yi; } cout << (s * 100 - total_cents) % 100 << endl; } ``` #### B. Odd and Even Pairs 此题目要求找到至少一对满足条件的索引:要么是一个偶数值的位置,或者是两个奇数值位置。程序会读入测试次数`t`以及每次测试中的数组长度`n`及其元素,并尝试找出符合条件的一对索引输出;如果没有这样的组合则返回-1[^3]。 ```cpp #include <cstdio> int main() { int t, n, num; scanf("%d", &t); while (t--) { int evenIndex = 0, oddIndex1 = 0, oddIndex2 = 0; scanf("%d", &n); for (int i = 1; i <= n; ++i) { scanf("%d", &num); if (num % 2 == 0 && !evenIndex) evenIndex = i; else if (num % 2 != 0) { if (!oddIndex1) oddIndex1 = i; else if (!oddIndex2) oddIndex2 = i; } if ((evenIndex || (oddIndex1 && oddIndex2))) break; } if (evenIndex) printf("1\n%d\n", evenIndex); else if (oddIndex1 && oddIndex2) printf("2\n%d %d\n", oddIndex1, oddIndex2); else printf("-1\n"); } return 0; } ``` 由于仅提供了前两道题的具体描述和解决方案,在这里无法继续给出完整的C至F题解答。通常情况下,每一道竞赛编程题都有其独特的挑战性和解决方法,建议查阅官方题解或社区讨论获取更多帮助。
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