网站与网站自动跳转代码,wordpress需要会php,南通学校网站建设,背景wordpressCF730F. Ber Patio
Solution
由于∑ai\sum{a_i}∑ai只有10510^5105#xff0c;即除掉原有的bbb#xff0c;代金券最多为10410^4104#xff0c;因此我们令f[i][j]f[i][j]f[i][j]表示到了第iii天#xff0c;靠现金获得的代金券共为jjj的最小现金和。
转移时枚举第i1i1i1…CF730F. Ber Patio
Solution
由于∑ai\sum{a_i}∑ai只有10510^5105即除掉原有的bbb代金券最多为10410^4104因此我们令f[i][j]f[i][j]f[i][j]表示到了第iii天靠现金获得的代金券共为jjj的最小现金和。
转移时枚举第i1i1i1天用kkk张代金券其他用现金的方法支付 upmin(f[i1][j(a[i1]−k)/10],f[i][j](a[i1]−k))upmin(f[i1][j(a[i1]-k)/10],f[i][j](a[i1]-k)) upmin(f[i1][j(a[i1]−k)/10],f[i][j](a[i1]−k)) 显然kkk最大不超过min(ai2,b−(s[i]−f[i][j])j)min(\frac{a_i}{2},b-(s[i]-f[i][j])j)min(2ai,b−(s[i]−f[i][j])j)能在3s3s3s内解决问题。
Code
#include vector
#include list
#include map
#include set
#include deque
#include queue
#include stack
#include bitset
#include algorithm
#include functional
#include numeric
#include utility
#include sstream
#include iostream
#include iomanip
#include cstdio
#include cmath
#include cstdlib
#include cctype
#include string
#include cstring
#include ctime
#include cassert
#include string.h
//#include unordered_set
//#include unordered_map
//#include bits/stdc.h#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i(a);i(b);i)
#define fi first
#define se second
#define int llusing namespace std;templatetypename Tinline bool upmin(T x,T y) { return yx?xy,1:0; }
templatetypename Tinline bool upmax(T x,T y) { return xy?xy,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pairint,int PR;
typedef vectorint VI;const lod eps1e-11;
const lod piacos(-1);
const int oo130;
const ll loo1ll62;
const int mods998244353;
const int MAXN300005;
const int INF0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f1,x0; char cgetchar();while (c0||c9) { if (c-) f-1; cgetchar(); }while (c0c9) { x(x3)(x1)(c^48); cgetchar(); }return x*f;
}
vectorint Ans;
int sum0,f[2][10005],a[5005],s[5005],frm[5005][10005];
signed main()
{int nread(),bread();for (int i1;in;i) s[i]s[i-1](a[i]read());for (int i0;i1;i)for (int j0;js[n]/10;j) f[i][j]INF;f[0][0]0;int nw0;for (int i0;in;i){nw^1;for (int j0;js[i]/10;j) f[nw][j]INF;for (int j0;js[i]/10;j)for (int k0;kmin(a[i1]/2,b-(s[i]-f[nw^1][j])j);k) if (upmin(f[nw][j(a[i1]-k)/10],f[nw^1][j]a[i1]-k)) frm[i1][j(a[i1]-k)/10]k;}int ansINF,Nw0;for (int i0;is[n]/10;i) if (upmin(ans,f[nw][i])) Nwi;printf(%lld\n,ans);for (int in;i1;i--) Ans.PB(frm[i][Nw]),NwNw-((a[i]-frm[i][Nw])/10);for (int in-1;i0;i--) printf(%lld ,Ans[i]);return 0;
}
