题目来源:福州大学acm代 码:fpcdq一入门熟悉ACM竞赛规则以及程序提交注意事项例题:Problem 1000 AB Problem Time Limit: 1000 mSec????Memory Limit : 32768 KBProblem DescriptionCalculate a b.InputThe input will consist of a series
CakeProblem Description一次生日Party可能有p人或者q人参加现准备有一个大蛋糕.问最少要将蛋糕切成多少块(每块大小不一定相等)才能使p人或者q人出席的任何一种情况都能平均将蛋糕分食.Input每行有两个数p和q.Output输出最少要将蛋糕切成多少块Sample Input2 3Sample Output4Hint将蛋糕切成大小分别为13131616的四块即满足要求
OJ上的一些水题(可用来练手和增加自信) (poj3299poj2159poj2739poj1083poj2262poj1503poj3006poj2255poj3094) 初期: 一.基本算法: (1)枚举. (poj1753poj2965) (2)贪心(poj1328poj2109poj2586) (3)递归和分治法. (4)递推. (5)构造法.
首先推荐大家一些非常简单的题特别适合没有算法基础的新手做(需要C语言基础) 1000 1001 1002 1003 1004 1005 1006 1007 1008 1012 1013 1017 1019 1023 1032 1045 1046 1047 1050 1061 1067 1068 1080 1083 1088 1095 1102 1132 1159 1163 1182 1183 12
TOC o 1-1 h z u l _Toc281217050 1001 Sum Problem PAGEREF _Toc281217050 h 2 l _Toc281217051 1089 AB for Input-Output Practice (I) PAGEREF _Toc281217051 h 4 l _Toc281217052 1090 AB for
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求体积include<>include<>define PI main(){ double x while(scanf(lfx)=EOF) { printf(.3lfn()) } return 0}求ab <> include<> define N 1005 char A[N]B[N]sum[N] int main() { int
(2)奇数阶魔方Time Limit: 20001000 MS (JavaOthers) Memory Limit: 6553632768 K (JavaOthers)Total Submission(s): 538 Accepted Submission(s): 311Problem Description一个 n 阶方阵的元素是12...n2它的每行每列和2条对角线上元素的
1002 A B Problem IITime Limit: 20001000 MS (JavaOthers)????Memory Limit: 6553632768 K (JavaOthers)Total Submission(s): 69615????Accepted Submission(s): 12678Problem DescriptionI have a very simple pr
2008年浙江省赛ACM题目一Accurately Say CocaCola Time limit: 1 Seconds?? Memory limit: 32768K?? Total Submit: 226?? Accepted Submit: 132?? In a party held by CocaColapany several students stand in a circ
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