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HDU 1061-Rightmost Digit(快速幂)
Rightmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32258    Accepted Submission(s): 12403 Problem Description Given a ...
分类:Web程序   时间:2014-09-03 09:43:26    阅读次数:189
Add Two Numbers
问题描述 You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers...
分类:其他好文   时间:2014-09-02 19:57:55    阅读次数:166
LeetCode 18 Letter Combinations of a Phone Number
Given a digit string, return all possible letter combinations that the number could represent. A mapping of digit to letters (just like on the telephone buttons) is given below. Input:Digit st...
分类:其他好文   时间:2014-09-01 21:15:03    阅读次数:223
Plus One
问题描述 Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list....
分类:其他好文   时间:2014-08-31 23:05:42    阅读次数:390
HNU Digit Sum (状态压缩)
Digit Sum Time Limit: 2000ms, Special Time Limit:5000ms, Memory Limit:65536KB Total submit users: 69, Accepted users: 61 Problem 12944 : No special judgement Problem description ...
分类:其他好文   时间:2014-08-30 20:29:30    阅读次数:245
UVa 1583 Digit Generator(数学)
?? 题意 如果a加上a所有数位上的数等于b时 a称为b的generator  求给定数的最小generator  给的数n是小于100,000的  考虑到所有数位和最大的数99,999的数位和也才45  因此我们只需要从n-45到n枚举就行了 #include #include using namespace std; int t, n, a, b, ans, l;...
分类:其他好文   时间:2014-08-30 09:59:39    阅读次数:192
HDU1061-Rightmost Digit(快速幂取模)
题目链接 题意:求n^n的个位数的值。 思路:快速幂求值 代码: #include #include #include #include #include using namespace std; typedef long long ll; ll a, b, n, m; struct mat{ ll s[2][2]; mat() {...
分类:Web程序   时间:2014-08-29 22:41:18    阅读次数:229
TOJ 1183 The Counting Problem 数位dp
http://acm.tju.edu.cn/toj/showp1183.html题意:从a写到b,问每个数字各写了几次。分析:昨天做了一道how many 0's,这题算是拓展。考虑从0..a。1到9是一样的,如果目前考虑的是digit,枚举每位,左边填0..x-1,则右边随便填(10^右边长度),...
分类:其他好文   时间:2014-08-29 19:40:48    阅读次数:211
HDU 1060 Leftmost Digit
题解:利用log,先计算答案的log值,n=n*log10(n),然后,最高位就是小数部分幂还原的整数部分。#include #include int T;double n;int main(){ scanf("%d",&T); while(T--){ scanf("%lf...
分类:其他好文   时间:2014-08-29 17:31:59    阅读次数:260
Plus One
Given a non-negative number represented as an array of digits, plus one to the number.The digits are stored such that the most significant digit is at...
分类:其他好文   时间:2014-08-25 14:42:24    阅读次数:138
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