C. Given Length and Sum of Digits...(思维题) 题意:给出两个数,m和s,m表示数字的位数(长度),s表示各个位上数字之和。输出符合条件的最大数和最小数。如果找不到符合条件的数就输出“-1 -1”。 题解:先考虑不符合条件的数,每个位上最大能放“9”,如果和大于位数x9,就找不到对应的值啦。当和大于0,
大数就是从最高位,能大就大;小数就是从最低位,能小就小,再处理下最高位为0的情况。 无结果无非一个sum太小,min全为0,一个sum太大,全为9还有剩 1publicclassMain {2publicstaticvoidmain(String[] args) {3Scanner io =newScanner(System.in);4intlen = io.nextInt(), sum =io.nextInt();5int...
#include <stack> #include #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits> #include <cstdlib> #include <cmath> #include #define maxn 1005 #define maxm 200005 #define eps 1e-10 #define mod 1000000007 #define INF 0x3f3f3...
#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<set>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>typedeflonglongll;usingnamespacestd;//freopen("","r",stdin);//freopen("D.out","w",stdout);int...
Primes with an average sum of digits, Compos. Math., 145 (2009), pp. 271-292.M. Drmota, C. Mauduit, J. Rivat. Primes with an average sum of digits, Compositio Math. 145, p.271-292, (2009).M. Drmota, C. Mauduit, and J. Rivat, Primes with an average sum of digits...
A. one B. one hundred C. one thousand D. one million 相关知识点: 试题来源: 解析 A One million 1000000, Adding, 1+9+0+0+0+0+0=1.一百万的数字之和是( ).A.1 B.100 C.1000 D.1000000一百万就等于1000000,所以1+0+0+0+0+0+0=1.故选A.反馈...
1926C-VladAndASumOfSumOfDigits.cpp 1926D-VladAndDivision.cpp 1927A-MakeItWhite.cpp 1927B-FollowingTheString.cpp 1927C-ChooseTheDifferentOnes.cpp 1927D-FindTheDifferentOnes.cpp 1927E-KleverPermutation.cpp 1928A-RectangleCutting.cpp 1928B-Equalize.cpp 1929A-SashaAndTheBeautifulArray.cpp 1929B-Sasha...
CObserve that 2019_(10)=5613_7.To maximize the sum of the digits, we want as many 65 as possible (since 6 is the highest value in base ), and this will occur with either of the numbers 4666_i or 5566_i.Thus, the answer is 4+6+6+6=(122..∼IronicNinja, edited by some peo...
A program to determine the sum of digits of a given non-negative integer number using a while loop is presented in Program. The program segment given below does the same thing using a do...while loop.
说明:双击或选中下面任意单词,将显示该词的音标、读音、翻译等;选中中文或多个词,将显示翻译。 1) sum of digits 位数和 例句>> 2) sum digit 和数位,和数数字 3) digital sum 位数码之和 1. In order to find the rules of the representation for integers under the factorial base,a kind ofdigital...