TYVJ1275
一道水题01背包,因为一时疏忽WA了4次RE1次
这题不一样的地方就是打每个小怪必须打死
也就是说每个小怪两下打不死要认为是三下。背包容量k,物品数量n
dp[j] = max(dp[j],dp[j-w[i]]+v[i]);
//667672 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define INF 0x3f3f3f3f using namespace std; const int maxn = 105; const int maxv = 25; int dp[maxv],w[maxn],v[maxn]; int main() { //freopen("in.txt","r",stdin); int n,m,k; cin>>n>>m>>k; for(int i = 1;i<=n;++i)cin>>w[i]; for(int i = 1;i<=n;++i)cin>>v[i]; if(!m || k==0){printf("0\n");return 0;} // for(int i = 1;i<=m;++i)dp[i]=-INF; // dp[0] = 0; memset(dp,0,sizeof(dp)); for(int i = 1;i<=n;++i) { if(w[i]%m)w[i] = w[i]/m+1; else w[i]/=m; for(int j = k;j>=w[i];--j) dp[j] = max(dp[j],dp[j-w[i]]+v[i]); } printf("%d\n",dp[k]); return 0; }
弱者究竟为何而战?!