C. Vlad and a Sum of Sum of Digits 分析 在给定的数据规模下,可以直接暴力计算出每个数的数位之和,求出前缀和后就可以在O(1)的时间复杂度下返回答案,因为数位拆分的时间复杂度为O(nlogn),所以总的时间复杂度为O(nlogn+t),其中t为查询次数。
Decrease the Sum of Digits (思维问题+构造)Codeforces Round #667 (Div. 3)』 昵称: unique_pursuit 园龄: 4年6个月 粉丝: 3 关注: 3 +加关注 < 2024年12月 > 日一二三四五六 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ...
You are given a positive integer n. Let S(x) be sum of digits in base 10 representation of xx, for example, S(123)=1+2+3=6, S(0)=0. Your task is to find two integers a,ba,b, such that 0≤a,b≤n, a+b=n and S(a)+S(b) is the largest possible among all such pairs...
Let f(x)f(x) be the sum of digits of a decimal number xx. Find the smallest non-negative integer xx such that f(x)+f(x+1)+⋯+f(x+k)=nf(x)+f(x+1)+⋯+f(x+k)=n. The first line contains one integer tt (1≤t≤1501≤t≤150)— the number of test cases. Each test...
Let $S(x)$ be sum of digits in base 10 representation of $x$, for example, $S(123) = 1 + 2 + 3 = 6$, $S(0) = 0$. Your task is to find two integers $a, b$, such that $0 \leq a, b \leq n$, $a + b = n$ and $S(a) + S(b)$ is the largest possible am...
CodeForces 489C (贪心) Given Length and Sum of Digits...,题意:找出m位且各个数位数字之和为s的最大和最小整数,不包括前导0(比如说003是非法的),但0是可以的。分析:这题是用贪心来做的,同样是m位数,前面的数字越大这个数就越大。所以写一个can(intm,ints)函数,来
sequences, and I had no luck. I've also tried rewriting it in many ways, etc, but I had no luck. I'm guessing you can just stimulate the process, but I'm pretty sure it would be very expensive (especially for large numbers where you need to obtain the kth digit of the number)...
Repeated Digit Sum of A^B By Md_Naimur_Rahman, history, 2 years ago, Here's the problem from LightOJ. It says to calculate sod(n) [sum of digits of n] while n >= 10. Here n = a^b, (0<= a,b <= 10^50,000 and a+b>0 )...
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提示: In the first sample the number already is one-digit — Herald can’t cast a spell. The second test contains number 10. After one casting of a spell it bec...