题目传送门 /* 这题交给队友做,做了一个多小时,全排列,RE数组越界,赛后发现读题读错了,囧! 贪心:先确定最高位的数字,然后用贪心的方法,越高位数字越大 注意:1. Both A and B will have same number of digits 两个数字位数相同 2. which is no larger than 10^6 不是大小,而是长度不超过1e6 */ #include <cstdio> #include <iostream> #include <al…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 513 Accepted Submission(s): 142 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1367    Accepted Submission(s): 327 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. Bu…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3902    Accepted Submission(s): 784 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. Bu…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 83    Accepted Submission(s): 16 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. But K…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number…

Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number when the sum of two digits exceeds 9. For example, when she calculates 4567+5789, she will get 9246, a…

DescriptionDoctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number when the sum of two digits exceeds 9. For example, when she calculates 4567+5789, she will get 9246, and for 12…

思路:刚开始想复杂了. 看解题报告后才知道这题挺简单的,看来还是要多训练啊!!! 单独处理首位的数字,不能为0.其他的就好处理了,从大到小依次找下去就可以了…… 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4726 题意:给出两个n位的数字,均无前缀0.重新排列两个数字中的各个数,重新排列后也无前缀0.得到的两个新数相加和最大.这里的相加是无进位的相加.输出结果.输出时不要输出前缀0. 思路:答案的第一位比较特殊,相加的两个数中不能有0.可以枚举这个答案中的第一个数字,查找是否存在两个数相加为这个数字.后面的道理一样,也是枚举. int a[2][10]; int n; char s[2][N]; int…

题目链接:http://acm.hdu.edu.cn/userstatus.php?user=zhsl 题意:给两个大数,他们之间的加法法则每位相加不进位.现在可以对两个大数的每位重新排序,但是首位不能为0,求两个数相加最大的和. 从最高位开始,考虑和最大贪心,每次从两个大数中找两位a和b,max=(a+b)%10的和最大.如果max有多种组成情况,对于a+b<10的所有情况,我们任意选择一组就可以了.对于存在a'+b'>10的组合情况,如果有a+b<10的组合情况,我们就选择a+b&l…

题目的意思是给你两个数字(多达10^6位) 做加法,但是有一点,没有进位(进位不算,相当于这一位相加后对10取模) 你可以任意排列两个数字中的每一位,但是不能是0开头. 现在题目要求以这种不进位的算法计算得到的最大值是多少? 看完题目就会知道,这个题目一定不是dp,或者说根本不是什么高端的算法,那是什么呢? 对,你没有猜错——贪心. 为什么可以用贪心呢?我们比较一个数,都是从高位的先比较,所以只要高位的大,这个数就大:于是我们要得到最大的和,就要先统计能够构成多少个9,多少个8……(有大的就先构…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 624    Accepted Submission(s): 178 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. Bu…

HDU 4716 A Computer Graphics Problem 水题.略 HDU 4717 The Moving Points 题目:给出n个点的起始位置以及速度矢量,问任意一个时刻使得最远的两点的距离最小. 分析:显然只有两点的话,答案满足三分性质.对于多个点,画个图分析一下,其实也满足三分性质.因此,先构造n*(n-1)/2个二次函数,于是三分枚举时间即可. #include <iostream> #include <cstdio> #include <cstr…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 605    Accepted Submission(s): 170 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. Bu…

Kia's Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3291    Accepted Submission(s): 703 Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. Bu…

HDU4716 A. A Computer Graphics Problem A题目描述 题意:输出手机剩余电量,保证给出的数是10的倍数. 题解:水题,按题意输出即可. 代码: #include <cstdio> #include <cstring> #include <algorithm> #define ll long long using namespace std; int main() { ,n,m; for (scanf("%d",&a…

pid=4726">http://acm.hdu.edu.cn/showproblem.php? pid=4726 大致题意:给两个长度小于10^6且相等的合法的正整数.你能够随意组合每一个数中的数字,但不能有前导零.两个数相加的规则如题,相加不进位. 问能够得到的A+B的最大值. 都看错题意了,一直以为数的大小是小于10^6,队友用了一个ms非常高端的函数对字符串全排列,枚举求最大值.结果WA到死.事实上是长度小于10^6,以后看题要细心再细心啊. .. 用数组记录每一个数中每一个数字的…

OpenCASCADE Curve Length Calculation eryar@163.com Abstract. The natural parametric equations of a curve are parametric equations that represent the curve in terms of a coordinate-independent parameter, generally arc length s, instead of an arbitray…

http://acm.hdu.edu.cn/showproblem.php?pid=1455 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=243 uva开头描述: 307 - Sticks Time limit: 3.000 seconds hduoj 开头描述: E - Sticks Time Limit:3000MS     Memo…

http://acm.hdu.edu.cn/showproblem.php?pid=4965 2014 Multi-University Training Contest 9 1006 Fast Matrix Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 238    Accepted Submission(…

1.现象 在vs2013中,按Ctrl + E + D格式化.cshtml代码,vs2013系统崩溃.报:inconsistent line count calculation in projection snapshot. 因为习惯性格式化代码,一格式化.cshtml就崩溃,需要重启vs2013,弄得我也不开心.因此这个问题必须解决. 2.原因 百度查原因,相关链接如下: http://stackoverflow.com/questions/917392/how-should-i-fix-sv…

http://acm.hdu.edu.cn/showproblem.php?pid=4712 Hamming Distance Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1610    Accepted Submission(s): 630 Problem Description (From wikipedia) For bina…

hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2005    Accepted Submission(s): 563 Problem Description Little Joh…

定义一个Strategy接口,其中定义一个方法,用于计算 using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Calculation { interface Interface1 { int calculate(int a, int b); } } 定义具体的算法类,实现Strategy接口,算法类中的算法各自不同:加减乘等 1,加法类 using Syst…

GLES2.0: Some device will give a warning on compling shaders(yet the compling will succeed), and the rendering result is incorrect with blink & artifacts. the problems is gradient calculation(interpolation) relies on the neighbor pixels, if the neigh…

List Variant: Configuration in Logistic General –> Retail Pricing –> Sales Price Calculation –> Define List Variant and -. The configuration is defining the list variant by defining the included fields and whether they can be changed in VKP5. Thi…

Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2272    Accepted Submission(s): 536 Problem Description Assume that f(0) = 1 and 0^0=1. f(n) = (n%10)^f(n/10) for all n bigger than ze…

Calculation 2 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is s…

Problem 1608 - Calculation Time Limit: 500MS   Memory Limit: 65536KB    Total Submit: 311  Accepted: 82  Special Judge: No Description Today, Alice got her math homework again! She had n integers, and she needed to divide them into several piles or o…