HDU 1003 最大连续子段和

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)
#### 问题描述
> 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.
#### 输入
> The first line of the input contains an integer T(1 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.

样例输入

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

样例输出

Case 1:

14 1 4

Case 2:

7 1 6

题意

求最大连续子段和，多解时求最左边那个。

题解

1、写wa了。。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=101010;

int arr[maxn],n;

int main() {
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        scf("%d",&n);
        for(int i=1;i<=n;i++) scf("%d",&arr[i]);

        int l=1,r=1,Ma=arr[1];
        for(int i=1;i<=n;i++){
            int tmp=arr[i];
            if(tmp>Ma){ Ma=tmp,l=r=i; }
            int ed=i+1;
			//这里其实强行把第i个塞进去了，这是错的。
            while(ed<=n&&tmp+arr[ed]>=0){
                tmp+=arr[ed];
                if(tmp>Ma){
                    Ma=tmp;
                    l=i,r=ed;
                }
                ed++;
            }
            i=ed-1;
        }

        prf("Case %d:\n",++kase);
        prf("%d %d %d\n",Ma,l,r);
        if(tc) prf("\n");
    }
    return 0;
}

2、改的很挫，终于过了。。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=101010;

int arr[maxn],n;

int main() {
//    freopen("data_in.txt","r",stdin);
//    freopen("data_out.txt","w",stdout);
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        scf("%d",&n);
        for(int i=1;i<=n;i++) scf("%d",&arr[i]);

        int l=0,r=0,Ma=-INF;
        for(int i=1;i<=n;i++){
            if(arr[i]<0){
                if(arr[i]>Ma){
                    Ma=arr[i];
                    l=r=i;
                }
                continue;
            }
            int tmp=arr[i];
            if(arr[i]>Ma){
                    Ma=arr[i];
                    l=r=i;
                }
            int ed=i+1;
            while(ed<=n&&tmp+arr[ed]>=0){
                tmp+=arr[ed];
                if(tmp>Ma){
                    Ma=tmp;
                    l=i,r=ed;
                }
                ed++;
            }
            i=ed-1;
        }

        prf("Case %d:\n",++kase);
        prf("%d %d %d\n",Ma,l,r);
        if(tc) prf("\n");
    }
    return 0;
}

//end-----------------------------------------------------------------------

3、来个正常点的思路：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=101010;

int arr[maxn],n;

int main() {
//    freopen("data_in.txt","r",stdin);
//    freopen("data_out.txt","w",stdout);
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        scf("%d",&n);
        for(int i=1;i<=n;i++) scf("%d",&arr[i]);

        int sum=0,l=1,r=1,tmp=1,Ma=-INF;

        for(int i=1;i<=n;i++){
            if(sum<0){
                sum=0; tmp=i;
            }
            sum+=arr[i];
            if(sum>Ma){
                Ma=sum;
                l=tmp,r=i;
            }
        }

        prf("Case %d:\n",++kase);
        prf("%d %d %d\n",Ma,l,r);

        if(tc) prf("\n");

    }
    return 0;
}

//end-----------------------------------------------------------------------