思路:

f[i][j]表示第i首歌音量为j是否可行(0/1)

初始f[0][bl]=1;

一层层往下推.最后取第n行最大的j 

 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 int n,bl,ml,f[51][1001],c[1001];
 5 int main(){
 6     memset(f,0,sizeof(f));
 7     scanf("%d %d %d",&n,&bl,&ml);
 8     for (int i=1;i<=n;i++)
 9         scanf("%d",&c[i]);
10     f[0][bl]=1;
11     for (int i=0;i<=n-1;i++)//i+1,第几首歌  
12         for (int j=0;j<=ml;j++)//音量
13             if (f[i][j]){//若当前音量已标记(可得到) 
14                 if (j-c[i+1]>=0)
15                     f[i+1][j-c[i+1]]=1;
16                 if (j+c[i+1]<=ml)
17                     f[i+1][j+c[i+1]]=1;
18             }
19     for (int i=ml;i>=1;i--)
20         if (f[n][i]){
21             printf("%d",i);return 0;
22         }
23     printf("-1");
24     return 0;
25 }
STD

 

posted on 2016-09-07 15:26  Absolutezero  阅读(215)  评论(0编辑  收藏  举报