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TYVJ1214

大水题,凑数
恰好装满的完全背包求最大最小价值
复习:完全背包从小到大枚举,恰好装满初始化要为无穷大正负分别考虑

 

 

 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #define INF  11111111 6 using namespace std; 7 const int maxn = 105; 8 const int maxv = 10005; 9 int dp[maxv],dp1[maxv];10 int a[maxn];11 int main()12 {13     int n,s;14     cin>>n>>s;15     for(int i = 1;i<=n;++i)cin>>a[i];16     for(int i = 1;i<=s;++i){dp[i] = -INF;dp1[i] = INF;}17     dp[0] = dp1[0] = 0;18     for(int i = 1;i<=n;++i)19         for(int j = a[i];j<=s;++j)20         {21             dp[j] = max(dp[j],dp[j-a[i]]+1);22             dp1[j] = min(dp1[j],dp1[j-a[i]]+1);23         }24     printf("%d\n%d\n",dp1[s],dp[s]);25     return 0;26 }

 

TYVJ1214