题目链接 隔了一年零三个月,重新刷URAL,这题挺麻烦的输出路径.输出路径挺扯的,乱写了写乱改改就A了...我本来想用很靠谱,记录每一条路径的,然后输出最小的,结果Tle,然后我使劲水水又过了一组,发现别人的题解..直接来了一次 就过了..我乱搞了搞,倒着记录最小的,然后倒着输出,就过了... #include <cstring> #include <cstdio> #include <string> #include <iostream> #include…

URAL 1658 思路: dp+记录路径 状态:dp[i][j]表示s1为i,s2为j的最小位数 初始状态:dp[0][0]=0 状态转移:dp[i][j]=min(dp[i-k][j-k*k]+1,dp[i][j])(0<=k<10) 在状态转移时用一个数组v[i][j]记录选的k 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define…

链接 这题卡了挺久了 昨天试着用类似dfs的方法直接TLE在第二组 看了下题解,,发现s1,s2的范围是个幌子..100位最大的s1900 s28100 觉得s1s2太大不敢开二维.. 这样就简单了 类似背包 dp[s1][s2]表示组成s2s2最少的位数 其实就是装进去多少个数字 正好把s1s2装满 把DP部分预处理之后放在外面  不然会超时 #include <iostream> #include<cstdio> #include<cstring> #include…

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…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…

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…

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…

ACdreamOJ 1154 Lowbit Sum (数位dp) ACM 题目地址:pid=1154" target="_blank" style="color:rgb(0,136,204); text-decoration:none">ACdreamOJ 1154 题意: long long ans = 0; for(int i = 1; i <= n; i ++) ans += lowbit(i) lowbit(i)的意思是将i转化成二进制…

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…

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…

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 ,…

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…

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(…

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…

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 …

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…

题意:AB两人分别拿一列n个数字,只能从左端或右端拿,不能同时从两端拿,可拿一个或多个,问在两人尽可能多拿的情况下,A最多比B多拿多少. 分析: 1.枚举先手拿的分界线,要么从左端拿,要么从右端拿,比较得最优解. 2.dp(i, j)---在区间(i, j)中A最多比B多拿多少. 3.tmp -= dfs(i + 1, r);//A拿了区间(l, i),B在剩下区间里尽可能拿最优 tmp是A拿的,dfs(i + 1, r)是B比A多拿的,假设dfs(i + 1, r)=y-x,y是B拿的,x是A…

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…

You are given two integers l l and r r (l≤r l≤r ). Your task is to calculate the sum of numbers from l l to r r (including l l and r r ) such that each number contains at most k k different digits, and print this sum modulo 998244353 998244353 . For…

1814: Ural 1519 Formula 1 Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 924  Solved: 351[Submit][Status][Discuss] Description  一个 m * n 的棋盘,有的格子存在障碍,求经过所有非障碍格子的哈密顿回路个数 Input The first line contains the integer numbers N and M (2 ≤ N, M ≤ 12). Each of…

测试样例之间输出空行,if(t>0) cout<<endl; 这样出最后一组测试样例之外,其它么每组测试样例之后都会输出一个空行. dp[i]表示以a[i]结尾的最大值,则:dp[i]=max(dp[i]+a[i],a[i]) 解释: 以a[i]结尾的最大值,要么是以a[i-1]为结尾的最大值+a[i],要么是a[i]自己本身,就是说,要么是连同之前的 构成一个多项的字串,要么自己单独作为一个字串,不会有其他的可能了. 状态规划的对状态的要求是:当前状态只与之前的状态有关,而且不影响下一…

题目链接 题意 : 定义Threeprime为它的任意连续3位上的数字,都构成一个3位的质数. 求对于一个n位数,存在多少个Threeprime数. 思路 : 记录[100, 999]范围内所有素数(标记的是该素数的每一位x1,x2,x3).然后从n = 4往后,定义dp[i][x2][x3],  i 表示到第 i 位时,第 i-1 位为 x2 , 第 i 位x3,此时所包含的情况数. dp[i][x2][x3] = dp[i][x2][x3] + dp[i-1][x1][x2]:最后求和sum…

C - Anniversary FireworkTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/B Description Denis has to prepare the Ural State University 90th anniversary firework. He bought n rockets and s…

Problem Description An equal sum partition of a sequence of numbers is a grouping of the numbers (in the same order as the original sequence) in such a way that each group has the same sum. For example, the sequence: 2 5 1 3 3 7 may be grouped as: (2…

Your task is to find the sum of all integer numbers lying between 1 and N inclusive. Input The input consists of a single integer N that is not greater than 10000 by it's absolute value. Output Write a single integer number that is the sum of all int…

这题和上次的通化邀请赛的那题一样,而且还是简化版本... 那题的题解      请戳这里 ... #include<cstdio> #include<algorithm> #include<cstring> #include<iostream> using namespace std; #define INF 0x3f3f3f3f int dp[105][105]; int a[105]; int sum,n; int pre_sum[105],next_s…

题目地址:Ural 1073 DP水题.也能够说是背包. #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #include <ctype.h> #include <queue> #include <map> #include…

1814: Ural 1519 Formula 1 Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 942  Solved: 356[Submit][Status][Discuss] Description Regardless of the fact, that Vologda could not get rights to hold the Winter Olympic games of 20**, it is well-known, that t…

题意:求一个区间内满足所有数位不同数字个数小于K的数字总和.比如:k=2   1,2,3所有数位的不同数字的个数为1满足,但是123数位上有三个不同的数字,即123不满足. 我们可以使用一个二进制的数字来记录某个数字是否已经出现,0为还没有出现,1表示该数字已经出现了.这里还需要注意前导零的干扰. #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include…

传送门--Codeforces 传送门--Atcoder 考虑逆序对的产生条件,是存在两个数\(i,j\)满足\(i < j,a_i > a_j\) 故设\(dp_{i,j}\)表示\(a_i>a_j\)的概率,每一次一个交换操作时\(O(n)\)地更新即可. AGC030D就在模意义下运算,最后就乘上\(2^Q\)就行了 看着好简单啊就是想不到 #include<iostream> #include<cstdio> #include<cstdlib>…