Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16488   Accepted: 11453 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will…

quicksum Given a string of digits, find the minimum number of additions required for the string to equal some target number. Each addition is the equivalent of inserting a plus sign somewhere into the string of digits. After all plus signs are insert…

Quicksum 时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte总提交:615            测试通过:256 描述 A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change…

Quicksum Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 102   Accepted Submission(s) : 33 Problem Description A checksum is an algorithm that scans a packet of data and returns a single number.…

POJ3094 Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18517   Accepted: 12712 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the check…

注意: for (int i = 1; i <= aaa.length(); i++) 其中是“ i <= ",注意等号. 原题: 2520.   Quicksum Time Limit: 0.5 Seconds   Memory Limit: 65536KTotal Runs: 2964   Accepted Runs: 1970 A checksum is an algorithm that scans a packet of data and returns a single…

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit cid=1006#status//H/0" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="font-family:Verdana,Arial,sans-serif; f…

Quicksum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3516 Accepted Submission(s): 2579 Problem Description A checksum is an algorithm that scans a packet of data and returns a single number. T…

http://poj.org/problem?id=3094 #include<iostream> #include <string> using namespace std; int main() { string str; getline(cin,str); while(str != "#") { ; ;i < str.length(); i++) { ; if(str[i] >= 'A' && str[i] <= 'Z')…

#include <stdio.h> #include <string.h> ]; int main() { ; int i,len; while(gets(word)) { sum = ; ] == '#') break; len = strlen(word); ; i < len; ++i) { if(word[i] != ' ') sum += (i+) * (word[i]-); } printf("%d\n",sum); } ; }…

[简要题意]:题意是非常easy. 看样能理解 [分析]:略. 读取字符串. // 200K 0Ms #include<iostream> using namespace std; int main() { char a[256]; while(1) { int sum = 0; gets(a); if(strcmp(a,"#")==0) break; int len = strlen(a); for(int i = 0;i<len;i++) { if(a[i] ==…

http://soj.me/1388 又一道字符串的水题.... #include <iostream> #include <cstring> using namespace std; int main() { char ch[260]; while(cin.getline(ch,260)) { if(ch[0] == '#') break; int ans = 0; int n = strlen(ch); for(int i=0;i < n;++i) { if(ch[i]…

T1.砍树(cuttree) 给出n棵树,要锯下m米木材,现在有一个高度h,h以上所有木头都砍下来,求满足m米的最小h 很简单的二分答案判断可行性. T2.快速求和(quicksum) 给出数字串s,要在其中加上一些加号,然后按正常表达式计算,求得到值n的最小添加数,如不能输-1. Dp(虽然std打的是dfs,但我是用dp AC的).F[i][j]表示前i个数中和为j的最小添加数,f[i][j]=min{f[i-k][j-num[k+1][j]]+1}.(好像和我程序里写的不太一样但是效果应该…

quicksum Queation: Given a string of digits, find the minimum number of additions required for the string to equal some target number. Each addition is the equivalent of inserting a plus sign somewhere into the string of digits. After all plus signs…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

描述 A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used for detecting transmission errors, validating document conten…

链接:第四次小练 A.POJ 3094   Quicksum 水题中的水题啊~ 直接上代码: #include <iostream> #include <cstdio> #include <cstring> using namespace std; ]; int main() { ) { memset(a,,sizeof(a)); gets(a); ]=='#') break; ,k; int n=strlen(a); ;i<n;i++) { k=a[i]-; ;…

Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13523   Accepted: 9407 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will…

各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1019 Grandpa's Other Estate 1034 Simple Arithmetics 1036 Complete the sequence! 1043 Maya Calendar 1054 Game Prediction 1057 Mileage Bank 1067 Rails 10…

HDU ACM-Steps RECORD Chapter 1 Section 1 暖手题 1.1.1 A+B for Input-Output Practice (I) #include <stdio.h> int main() { int a,b; while(scanf("%d %d",&a,&b)==2) printf("%d\n",a+b); return 0; } 1.1.2 A+B for Input-Output Pract…

Quicksum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3401 Accepted Submission(s): 2095   Problem Description A checksum is an algorithm that scans a packet of data and returns a single number.…

2992.357000 1000 A+B Problem1214.840000 1002 487-32791070.603000 1004 Financial Management880.192000 1003 Hangover792.762000 1001 Exponentiation752.486000 1006 Biorhythms705.902000 1005 I Think I Need a Houseboat686.540000 1011 Sticks647.566000 1007…

以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…

1. POJ 3299 Humidex 链接: http://poj.org/problem?id=3299 这道题是已知H,D,T三者的运算关系,然后告诉你其中两个.求另一个. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; int main () { char c,h; double T,D,H; while(cin>…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

Quicksum Time Limit: 2 Seconds      Memory Limit: 65536 KB A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used for d…

Gurobi使用教程 1.Gurobi使用的一般框架 from gurobipy import * try: m=Model('modelname') except GurobiError: print('Error reported') 2.Gurobi读取数据求解 假设给定三个文件,分别是categories.txt,foodcost.txt以及nutritionvalues.txt,分别存放每天营养摄取的上限和下限,每种食物的价格以及每种食物所含的营养成分,其中categories.txt…

1.quicksum好像不支持嵌套 最好还是尽可能多的使用一些中间变量来表达. 另外,quicksum()返回的是表达式而不是var,像addGenConstrMax(resvar, [var, var, var, ...])这样的函数中,quicksum也不能直接传给var, 对此只能再创建中间变量来保存quicksum()的求解结果,然后把这个中间变量传给addGenConstrMax. 其实多使用些中间变量也容易理清思路,不易出错. 2.上述所说的addGenConstrMax是gurob…

场景: 使用gurobi求解优化问题时,遇到quicksum()函数用法如下: quicksum(mu[i] for i in range(n)) 读着很流畅而且好像并没什么问题欸,但 mu[i] for i in range(n) 返回的又是什么? 看了下quicksum()函数的介绍: def quicksum(p_list): # real signature unknown; restored from __doc__ """ ROUTINE: quicksum(li…

Quicksum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8353    Accepted Submission(s): 5614 Problem Description A checksum is an algorithm that scans a packet of data and returns a single num…