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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }