|
Problem Description Given 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 sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input 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). Output 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. Sample Input 2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5 Sample Output Case 1: 14 1 4 Case 2: 7 1 6 源自http://acm.hdu.edu.cn/ 1003题 欢迎大家踊跃参与。。。 |
[技术| 编程·课件·Linux] ACM题天天练 4 Max Sum
sunhongbo
· 发布于 2012-08-23 12:05
· 6121 次阅读
转载文章时务必注明原作者及原始链接，并注明「发表于 软院网 RuanYuan.Net 」，并不得对作品进行修改。
|
本帖最后由 hslx111 于 2012-8-23 13:36 编辑 [C++] 纯文本查看 复制代码
//hud 1003
#include <iostream>
using namespace std;
int main()
{
int i,j,T,N;
int num=1;
cin>>T;
while(T--)
{
int start,end,pos,sum,now,temp;
cin>>N>>temp;
now=sum=temp;
start=end=pos=1;
for(i=2;i<=N;i++)
{
cin>>temp;
if(now+temp<temp)
now=temp,pos=i;
else
now+=temp;
if(now>sum)
sum=now,start=pos,end=i;
}
cout<<"Case "<<num<<":"<<endl;
cout<<sum<<" "<<start<<" "<<end<<endl;
if(T!=0) cout<<endl;
num++;
}
return 0;
}
好长时间不写,手都生了啊.... |
点评
评分
|
本帖最后由 yel_hb 于 2012-8-23 13:58 编辑 ...看了楼上的贴自己代码有点心虚的说...  [C++] 纯文本查看 复制代码 #include <iostream>
using namespace std;
int main()
{
int t,n,i,k,max,sum,x1,x2,y1;
int *list;
cin >> t;
k = 1;
while(t--)
{
cin >> n;
list = new int[n];
for(i = 0; i < n; ++i)
{
cin >> list[i];
}
x1 = 0;
y1 = 0;
x2 = 0;
max = list[x1];
sum = list[x1];
for(i = x1 + 1; i < n; ++i)
{
if(sum >= 0)
{
sum += list[i];
}
else
{
x2 = i;
sum = list[i];
}
if(max < sum)
{
max = sum;
x1 = x2;
y1 = i;
}
}
cout << "Case " << k++ << ":" << endl;
cout << max << " " << (x1 + 1) << " " << (y1 + 1) << endl;
if(t != 0)
{
cout << endl;
}
}
return 0;
}
|
评分
|
本帖最后由 justcx 于 2012-8-23 14:41 编辑 [C++] 纯文本查看 复制代码 #include<iostream>
#include<iostream>
#define N 100000
static st,ed;
int b[N];
int MaxSumDP(int *a, int n);
using namespace std;
int main()
{
int n1,n2;
int a[N];
cin>>n1;
for(int j=0;j<n1;j++)
{
cin>>n2;
for(int i=0;i<n2;i++)
{
cin>>a[i];
}
int M=MaxSumDP(a,n2);
cout<<"Case :"<<j+1<<endl;
cout<<M<<' '<<st<<' '<<ed<<endl;
cout<<endl;
}
return 0;
}
int MaxSumDP(int *a, int n)
{
int i, maxSum = 0;
int count=0;
if(a[0] >=0)
{
maxSum=b[0]=a[0];
st=1;
ed=1;
count++;
}
for(i=1; i<n; ++i)
{
b = (b[i-1]+a > 0) ? (b[i-1]+a) : 0;
if(b[i-1]==0)
count++;
while(count==0)
{
if(b[i-1]==0&&b>0)
{
st=i+1;
}
}
if(maxSum < b[i])
{
maxSum = b[i];
ed=i+1;
}
}
return maxSum;
}
|
评分
谢谢支持谢谢参与，真棒。。。
。。。|
无语，调试没问题，谁能告诉我为何一直wa？ [Java] 纯文本查看 复制代码 package com.slk.MaxSumSubsequence;
import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int t, n;
int pos = 1;
int spos = 1;
int epos = 1;
int caseNum = 1;
t = Integer.parseInt(sc.nextLine());
while (t-- > 0) {
n = Integer.parseInt(sc.nextLine());
int currentInput = Integer.parseInt(sc.nextLine());
int tempSum = currentInput;
int maxSum = tempSum;
for (int i = 1; i < n; i++) {
currentInput = Integer.parseInt(sc.nextLine());
if (currentInput + tempSum < currentInput) {
tempSum = currentInput;
pos = i + 1;
} else {
tempSum += currentInput;
}
if (tempSum > maxSum) {
maxSum = tempSum;
spos = pos;
epos = i + 1;
}
}
System.out.println("Case " + caseNum++ + ":");
System.out.println(maxSum + " " + spos + " " + epos + "\n");
}
sc.close();
}
}
|
本帖最后由 maxOrder石 于 2012-8-23 17:55 编辑 hslx111 发表于 2012-8-23 13:34 不用检测T的值和N的值吗，我就纳闷了，每次写完结果正确就是提交不了，求改，谢谢 #include<stdio.h> int main() { int T; scanf("%d",&T); if(T<1||T>20) { return 0; } int N; int i; int j; int s,MaxSum[20]={0}; int starP[20],endP[20]={0}; for(j=0;j<T;j++) { scanf("%d",&N); if(N<1||N>100000) return 0; starP[j]=1; int *number=(int *)malloc(sizeof(int)*N); for(i=0;i<N;i++) { scanf("%d",number+i); } s=0; for(i=0;i<N;i++) { s=s+*(number+i); if(MaxSum[j]<s) { MaxSum[j]=s; endP[j]=i+1; } } } for(i=0;i<T;i++) { printf("Case %d:\n",i+1); printf("%d %d %d\n\n",MaxSum,starP,endP); } return 0; } 字体变了，最后的printf（是MaxSum中括号i,后面都是一样的额） |
