HDU 1003 Max Sum(dp，最大連續子序列和)

標籤：acm   hdu   動態規劃   dp   

Max SumProblem 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 sum in this sequence is 6 + (-1) + 5 + 4 = 14.
 
InputThe 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).
 OutputFor 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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 Sample Output

Case 1:14 1 4Case 2:7 1 6

題意  求n個數位最大連續和

DP的入門題目  令d[i]表示以第i個數a為右端的最大連續子序列和   那麼很容易得出轉移方程  d[i]=max(d[i-1]+a,a)

很顯然  當第i個數比以第i-1個數為右端的最大和加上第i個數還大的時候  以第i個數為右端的最大和就是第i個數自己了   同時更新左端為自己

#include<cstdio>#include<cstring>using namespace std;const int N = 100005;int main(){    int a, cas, ans, l, le, ri, n, d[N];    scanf ("%d", &cas);    for (int k = 1; k <= cas; ++k)    {        memset (d, 0x8f, sizeof (d));        ans = d[0];        scanf ("%d", &n);        for (int i = 1; i <= n; ++i)        {            scanf ("%d", &a);            if (d[i - 1] + a < a)                d[i] = a, l = i;            else                d[i] = d[i - 1] + a;            if (d[i] > ans)                ans = d[i], le = l, ri = i;        }        if (k > 1) printf ("\n");        printf ("Case %d:\n%d %d %d\n", k, ans, le, ri);    }    return 0;}