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<=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). #### 输出 > 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

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

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 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

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

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 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

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

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 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-----------------------------------------------------------------------