foj2013

weixin_34319374

于 2019-06-13 07:57:15 发布

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原文链接：https://my.oschina.net/garyun/blog/602972

本文深入探讨了一个算法问题：如何找出长度不小于M的子串中，元素和最大的那个子串。通过详细的代码示例，本文展示了如何使用动态规划的方法解决这一问题，提供了一种高效且易于理解的解决方案。

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 // Problem 2013 A short problem  
/* Accept: 158 Submit: 452
Time Limit: 1000 mSec Memory Limit : 32768 KB
 Problem Description
The description of this problem is very short. Now give you a string(length N), and ask you the max sum of the substring which the length can't small than M.
Input
The first line is one integer T(T≤20) indicates the number of the test cases. Then for every case, the first line is two integer N(1≤N≤1000000) and M(1≤M≤N).
Then one line contains N integer indicate the number. All the number is between -10000 and 10000.
Output
Output one line with an integer.
Sample Input
2
5 1
1 -2 -2 -2 1
5 2
1 -2 -2 -2 1
 Sample Output
1
-1
 */ 
#include " iostream " 
 using   namespace  std;
 const   int  M = 1000005 ;
 int  t,n,m;
 int  i,j;
 int  a[M];
 int  sum[M];
 int  f[M];
 int  cmp( int  a, int  b)
{
 return  a > b ? a:b;
}
 int  main()
{
 scanf( " %d " , & t);
 while (t -- )
 {
 scanf( " %d %d " , & n, & m);
 for (i = 1 ;i <= n;i ++ ) scanf( " %d " , & a[i]);
 int  i,Max =- 0x7fffffff ;
 sum[ 0 ] = 0 ;
 for (i = 1 ;i <= n;i ++ ) sum[i] = sum[i - 1 ] + a[i];
 f[m - 1 ] = sum[m - 1 ];

 for (i = m; i <= n; i ++ ) 
 {
 f[i] = cmp( f[i - 1 ] + a[i], sum[i] - sum[i - m]);
 if (f[i] > Max) Max = f[i];
 }
 printf( " %d\n " ,Max);
 }
 return   0 ;
}