随笔分类 - ACM-UVA
摘要:L -Minimum Sum LCMTime Limit:3000MSMemory Limit:0KB64bit IO Format:%lld & %lluSubmitStatusPracticeUVA 10791题意:输入正整数n,找至少两个数,使得他们的LCM为n且要输出最小的和;思路:既然LC...
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摘要:I -Teams Time Limit:1000MSMemory Limit:0KB64bit IO Format:%lld & %lluSubmitStatusPracticeUVA 11609题意:有n个人,选多个人参加比赛,其中一个是队长,队长不同其他选手...
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摘要:K - Integer GameTime Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %lluSubmit Status Practice UVA 11489题意:在n中取数字,使剩下的数是3的倍数,不能取则失败。思路:如果能使当前数为3...
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摘要:Add AgainSummation of sequence of integers is always a common problem in Computer Science. Rather than computing blindly, some intelligent techniques ...
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摘要:Again Prime? No time.Input: standard inputOutput: standard outputTime Limit: 1 secondThe problem statement is very easy. Given a number n you have to ...
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摘要:题意:Morley定理,求D、E、F的坐标 思路:没什么算法,就是几何的应用。注意旋转角就好了。转载请注明出处:寻找&星空の孩子题目链接:UVA11178 1 #include 2 #include 3 #define PI acos(-1.0) 4 using namespace std; 5 ...
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摘要:Power of MatrixTime Limit:3000MSMemory Limit:0KB64bit IO Format:%lld & %lluSubmitStatusPracticeUVA 11149Appoint description:DescriptionProblem B : Pow...
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摘要:Problem EContemplation! AlgebraInput:Standard InputOutput:Standard OutputTime Limit:1 SecondGiven the value ofa+bandabyou will have to find the value ...
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摘要:>>什么是01背包>题目链接#includeusing namespace std;int main(){ int N,V,T; int w[1005]; int c[1005]; int i,j,k; scanf("%d",&T); while(T--) ...
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摘要:11078 - Open Credit SystemTime limit: 3.000 secondsProblem EOpen Credit SystemInput:Standard InputOutput:Standard OutputIn an open credit system, the ...
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摘要:Problem AHashmat the brave warriorInput:standard inputOutput:standard outputHashmat is a brave warrior who with his group of young soldiers moves from...
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