CF349 B. Color the Fence
阿新 • • 發佈:2021-07-01
題目傳送門:https://codeforces.com/problemset/problem/349/B
題目大意:
給你總顏料數\(v\),再給你9個數\(a_{1...9}\),\(a_i\)表示畫\(i\)的顏料消耗,求所能畫出的最大的數
要考慮畫的數最大,首先得看位數,位數 \(len\) 可以用 $\lfloor \frac{v}{Min}\rfloor $確定,而 \(Min=\min\{a_i\}\)
得到位數後,我們從高位,從大數開始
如果 \(\lfloor\frac{v-a_i}{Min}\rfloor=len-1\) ,則說明該位可填,否則不行
/*program from Wolfycz*/ #include<map> #include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair<int,int> #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } template<typename T>inline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } template<typename T>inline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=10; int A[N+10]; int main(){ // freopen(".in","r",stdin); // freopen(".out","w",stdout); int v=read(0),Min=int_inf; for (int i=1;i<10;i++) A[i]=read(0),Min=min(Min,A[i]); if (v<Min){ printf("-1\n"); return 0; } int Cnt=v/Min; while (Cnt--){ for (int i=9;i;i--){ if (v>=A[i]&&(v-A[i])/Min==Cnt){ printf("%d",i); v-=A[i]; break; } } } putchar('\n'); return 0; }