杭电ACM题1003

杭电题1003

Problem Description
Given a sequence a[1],a[2],a[3]......a
, your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

[align=left]Input[/align]
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and
1000).

[align=left]Output[/align]
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end
position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

[align=left]Sample Input[/align]

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

[align=left]Sample Output[/align]

Case 1:
14 1 4

Case 2:
7 1 6

这题跟我有仇哈。。。。
[code]#include "iostream"
using namespace std ;

int main()
{
int a[1003];//存储输入的一些数
int str[1003];//存储依次计算得到的和
int start[1003];//存储序列和种最大值的最先开始元素序号
int t,n;
int num=1;
cin>>t;
while (t--)//case的次数
{
cin>>n;//输入n个数
for(int i=1;i<=n;i++){
cin>>a[i];
}
str[1]=a[1];
start[1]=1;
for (int j =2;j<= n;j++)
{
if (str[j-1]>=0)//元素数和大于0则继续看下一位
{
str[j]=str[j-1]+a[j];
start[j]=start[j-1];
}else{//小于0从写一个元素记录开始
str[j]=a[j];
start[j]=j;
}
}
int max=str[1];
int end=1;//随便初始化一下
for ( int k = 0; k<=n; k++)//找到序列和中的最大值
{
if (str[k]>max)
{
max=str[k];
end=k;
}

}
cout<<"Case"<<" "<<num<<":"<<endl;
num++;
cout<<max<<" "<<start[end]<<" "<<end;
if (t) cout<<endl;

}
return 0;
}

<span style="color:#000066;">跟着学习了，原来这样才能通过杭电的编辑器</span>

#include "iostream"
using namespace std ;

int main(){
int r = 0,l = 0,i= 0 ,j = 0,num =0,n,t=0;
int sum = 0,max = 0;
int *a;
cin>>n;
while(n--){
cin>>num;
a = (int *)calloc(num,sizeof(int));
for(i = 0; i < num ;i ++)
cin>>a[i];
for( l = 0,r = 0,sum = 0,max = a[0],i = 0;i <num ;i ++)
{
for(sum = 0,j = i ;j <num ;j ++)
{
sum += a[j];
if(sum > max){
max = sum ;
l = i;
r = j;
}
if(sum < 0){
i = j;
sum =0;
break;
}
}
}
t++;
cout<<"Case"<<" "<<t<<":"<<endl;
cout<<max<<" "<<l+1<<" "<<r+1<<endl;
if (n) cout<<endl;
}
system("pause");
return 0;
}

[/code]