MemSQL Start[c]UP 2.0 - Round 1 D. Washer, Dryer, FoldercO
2014-08-03 16:31
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Codeforces MemSQL Start[c]UP 2.0 - Round 1 D. Washer, Dryer, FoldercO
题解:
背包问题
把个数当做容量,值为时间
题解:
背包问题
把个数当做容量,值为时间
#include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include <complex> #include <fstream> #include <cassert> #include <cstdio> #include <bitset> #include <vector> #include <deque> #include <queue> #include <stack> #include <ctime> #include <set> #include <map> #include <cmath> #define eps 1e-9 #define INF 0x3f3f3f3f using namespace std; typedef long long ll; typedef long double ld; typedef pair<ll, ll> pll; typedef complex<ld> point; typedef pair<int, int> pii; typedef pair<pii, int> piii; template<class T> inline bool read(T &n) { T x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true; } template <class T> inline void write(T n) { if(n < 0) { putchar('-'); n = -n; } int len = 0,data[20]; while(n) { data[len++] = n%10; n /= 10; } if(!len) data[len++] = 0; while(len--) putchar(data[len]+48); } //----------------------------------- const int MAXN=10010; int k,n1,n2,n3,t1,t2,t3; int t[MAXN]; int main() { read(k);read(n1);read(n2);read(n3);read(t1);read(t2);read(t3); for(int i=0;i<k;i++) { if(i>=n1)t[i]=max(t[i],t[i-n1]+t1); if(i>=n2)t[i]=max(t[i],t[i-n2]+t2); if(i>=n3)t[i]=max(t[i],t[i-n3]+t3); } printf("%d\n",t[k-1]+t1+t2+t3); return 0; }
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