Given Length and Sum of Digits... 题目链接: http://acm.hust.edu.cn/vjudge/contest/121332#problem/F Description You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m…

C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the la…

题目链接:http://codeforces.com/problemset/problem/489/C 题目意思:给出 m 和 s,需要构造最大和最小的数.满足长度都为 m,每一位的数字之和等于 s.如果构造不出来,输出 -1 -1.否则输出最小和最大且符合条件的数. 想了两个多小时,发现想错了方向...... /****************************************** 首先把不能得到最小数和最大数的情况揪出来. 第二组测试数据 3 0 有提示,s = 0 且 m >…

m位长度,S为各位的和 利用贪心的思想逐位判断过去即可 详细的注释已经在代码里啦~ //#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <cstring> #include <cmath> #include <stack> #include <queu…

http://codeforces.com/problemset/problem/489/C C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer …

C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the la…

C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the la…

You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m and sum of digits s. The required numbers should be non-negative integers written in the decimal base with…

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(int m, int s)函数,来判断是否存在一个m位数其各位数字之和为s 这里先不考虑前导0的事,代码看起来可能是这个样子的: bool can(int m, int s) { && s <= m*); } 比如我们现在要求满足要求的最小整数,从最左边的数开始从0到9开始试,如果后面的…

题意:给你一个正整数\(n\),每次可以对\(n\)加一,问最少操作多少次是的\(n\)的所有位数之和不大于\(s\). 题解:\(n\)的某个位置上的数进位,意味这后面的位置都可以被更新为\(0\),所以我们从高位往低位记录一个\(sum\),然后根据情况判断即可. 代码: int t; int s; ll n; char str[N]; int main() { //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); t=read(); w…

#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <t…

Maximum Sum of Digits You are given a positive integer n. Let S(x)S(x) be sum of digits in base 10 representation of xx , for example, S(123)=1+2+3=6S(123)=1+2+3=6 , S(0)=0S(0)=0 . Your task is to find two integers a,ba,b , such that 0≤a,b≤n0≤a,b≤n ,…

Description You are given a positive integer nn. 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…

Sum of Digits / Digital Root In this kata, you must create a digital root function. A digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has two digits, continue reducing in this w…

Sum of Digits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 810    Accepted Submission(s): 220 Problem Description Petka thought of a positive integer n and reported to Chapayev the sum of its…

D. Roman Digits time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers…

B. Maximum Sum of Digits time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output You are given a positive integer nn. Let S(x)S(x) be sum of digits in base 10 representation of xx, for example, S(…

链接 Codeforces 703D Mishka and Interesting sum 题意 求区间内数字出现次数为偶数的数的异或和 思路 区间内直接异或的话得到的是出现次数为奇数的异或和,要得到偶数的需要把区间内出现过的数字不重复的再异或一遍.离线按右端点排序,每次处理一个区间时,如果该数字出现过,则在树状数组中把这个数删去,再重新再该位置加到树状数组中. 代码…

Educational Codeforces Round 53 E. Segment Sum 题意: 问[L,R]区间内有多少个数满足:其由不超过k种数字构成. 思路: 数位DP裸题,也比较好想.由于没考虑到前导0,卡了很久.但最惨的是,由于每次求和的时候需要用到10的pos次幂,我是用提前算好的10的最高次幂,然后每次除以10往下传参.但我手贱取模了,导致每次除以10之后答案就不同余了,这个NC细节错误卡了我一小时才发现. 代码: #include<iostream> #include<…

题目如下: Given an integer number n, return the difference between the product of its digits and the sum of its digits. Example 1: Input: n = 234 Output: 15 Explanation: Product of digits = 2 * 3 * 4 = 24 Sum of digits = 2 + 3 + 4 = 9 Result = 24 - 9 = 1…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 遍历 日期 题目地址:https://leetcode-cn.com/problems/sum-of-digits-in-the-minimum-number/ 题目描述 Given an array A of positive integers, let S be the sum of the digits of the minimal elemen…

A. Tricky Sum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/598/problem/A Description In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.…

[题目链接] http://codeforces.com/contest/703/problem/D [题目大意] 给出一个数列以及m个询问,每个询问要求求出[L,R]区间内出现次数为偶数的数的异或和. [题解] 显然,我们很容易求出区间内出现次数为奇数的数的异或和,那么如果我们可以求出区间内出现的所有数的异或和,那么将两者异或就可以得到要求的东西. 我们记一个数字上一次出现的位置为pre,对于[L,R]中的数,如果其pre是小于L的,那么它肯定是第一次在这个区间出现,所以现在问题就转化为求[L…

F. The Sum of the k-th Powers 题目连接: http://www.codeforces.com/contest/622/problem/F Description There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find the value of the sum modulo 109 + 7 (so you shoul…

Instructions In this kata, you must create a digital root function. A digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has more than one digit, continue reducing in this way unti…

链接: https://codeforces.com/contest/1228/problem/A 题意: You have two integers l and r. Find an integer x which satisfies the conditions below: l≤x≤r. All digits of x are different. If there are multiple answers, print any of them. 思路: 水题. 代码: #include…

题目: http://codeforces.com/contest/608/problem/B 字符串a和字符串b进行比较,以题目中的第一个样例为例,我刚开始的想法是拿01与00.01.11.11从左到右挨个比较,希望能找到一些规律,结果并没有... 其实,如果我们能从整个比较过程来看这个问题,整个过程就没有那么难.题目要求的东西,其实就是a字符串和b字符串子串每次比较的不同的个数的总和,当我们像上面那个思路,拿a字符串每次移动一位,和b进行比较的时候,从整个过程来看,就相当于a的每一个元素从前…

The Sum of the k-th Powers There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find the value of the sum modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).…

异或运算性质,离线操作,区间求异或和. 直接求区间出现偶数次数的异或和并不好算,需要计算反面. 首先,很容易求解区间异或和,记为$P$. 例如下面这个序列,$P = A[1]xorA[2]xorA[3]......xorA[15]$ $1$,$1$,$1$,$2$,$2$,$3$,$3$,$3$,$4$,$4$,$5$,$5$,$6$,$7$,$7$. 出现偶数次数的异或和记为$Q$,那么$Q = 2xor4xor5xor7$. 我们记$F=PxorQ$,如果知道$F$,那么就能计算出$Q$.所…

题目传送门 题目大意:给出n个数字,m次区间询问,每一次区间询问都是询问 l 到 r 之间出现次数为偶数的数 的亦或和. 思路:偶数个相同数字亦或得到0,奇数个亦或得到本身,那么如果把一段区间暴力亦或,得到的其实就是出现次数为奇数的数字的亦或和,所以我们希望这段区间内的所有数字出现次数都+1,使奇偶性互换. 我们先处理出前缀的亦或和,这样可以得到次数为奇数的亦或和. 接下来的问题就是要改变一段区间的奇偶性了,也就是说,这个问题其实就转化成了如何求一段区间出现的所有数字(无重复). 这里我学到的是…