摘要：Max SumTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 59774Accepted Submission(s): 13593Problem DescriptionGiven a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max
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文章分类 - 【1-03】【HDUOJ】
学习C++,C,Java。研究数据结构，算法。
提高编码内功。
摘要：#include<iostream>using namespace std;#define Base 10#define MaxLen 1000struct BigInt { int len; int data[MaxLen]; BigInt() :len(0) {} BigInt(const BigInt& s) :len(s.len){ memcpy(this->data,s.data,len*sizeof(*data)); } BigInt(int s) :len(0){ for(;s>0;s=s/Base) data[len++]=s%Base; } B
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摘要：#include<iostream>using namespace std;int main(){int a,b;int sum=0;cin>>a>>b;for(int i=1;i<=b;i++){sum=sum+i;}return 1;}#include<iostream>using namespace std;int main(){int a,b;int sum=0;cin>>a>>b;for(;a<=b;a++){sum=sum+a;}count<<a<<endl;cout<
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摘要：C++#include<iostream>using namespace std;int main(){int a,b;while(cin>>a>>b)cout<<a+b<<endl;return 1;}C#include<stdio.h>#define P 3.1415927main(){ int a,b; while(scanf("%d%d",&a,&b)==2) printf("%d\n",a+b); return 0;}Javaimport java.util
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