BZOJ2748

BZOJ2748

• 题目

  BZOJ2748

• 分析

  考虑 $DP$

  $f[i][j]$ 表示音量为 $j$ 可以由前 $i$ 次增加或减少构成

  $$f[i][j]=1(f[i-1][j+a[i]]=1||f[i-1][j-a[i]]=1)$$

  if (j + a[i] <= ml && f[i - 1][j + a[i]]) // 前i次变换后减a[i]得到f[i][j]
  		f[i][j] = 1;
  if (j - a[i] >= 0 && f[i - 1][j - a[i]])//前i次变换后加a[i]得到f[i][j]
  	    f[i][j] = 1;
  

• 代码

  const int N = 55;
  int a[N];
  int f[N][1005];
  int main ()
  {
         
  	//freopen("input.in", "r", stdin);
  	//freopen("test.out", "w", stdout);
  	int n, bl, ml;
  	read(n);
  	read(bl);
  	read(ml);
  	for (int i = 1; i <= n; i++) read(a[i]);
  	memset(f, 0, sizeof(f));
  	f[0][bl] = 1;
  	for (int i = 1; i <= n; i++)
  	{
         
  		for (int j = 0; j <= ml; j++)
  		{
         
  			if (j + a[i] <= ml && f[i - 1][j + a[i]])
  				f[i][j] = 1;
  			if (j - a[i] >= 0 && f[i - 1][j - a[i]])
  				f[i][j] = 1;
  		}
  	}
  	int ans = 0;
  	for (int i = ml; i >= 0; i--)
  		if (f[n][i]) {
         ans = i; break;}
  	if (ans) cout << ans << endl;
  	else cout << -1 << endl;
  	return 0;
  }
  

• 题型

  $DP$