Codeforces 420
A
Problem description
有k个物品和v个隔板,每个箱子最多分成b个空格,每个空格里最多放a个物品,求放下所有物品最少要多少箱子.
Data Limit2 ≤ k ≤ 1000; 1 ≤ a, b, v ≤ 1000 Time Limit: 1s
Solution
虽说可以O(1)出解,但比较麻烦容易错,反正数据小,为了正确性就模拟一下,考虑每个箱子,隔成最多的空格然后放最多的物品.
Code
#include<cstdio> int n,k,t,a,b,v,ans=0; int min(int x,int y) { if (x>y)return y;else return x; } int main() { scanf("%d%d%d%d",&k,&a,&b,&v); while (a>0) { ans++; t=min(b,k-1);b-=t; if(a<=(t+1)*v) { printf("%d",ans); return 0; } a-=(t+1)*v; } return 0; }
B
Problem description
给出n和n棵树的高度a以及等差k,修建最少的树使n棵树的高度为等差为k的等差数列(修建可以使树增高,但不能修成负数)
Data Limit：1 ≤ n, k ≤ 1000 1 ≤ ai ≤ 1000 Time Limit: 1s
Solution
由于是等差数列,所以枚举第一棵树的高度就可以啦,还可以避免出现负数.
Code
#include<cstdio> int n,i,j,a[10000],k,t=0,min=1e8,mini; int main() { scanf("%d%d",&n,&k); for (i=1;i<=n;i++)scanf("%d",&a[i]); for (i=1;i<=10000;i++) { t=0; for (j=1;j<=n;j++) if (i+(j-1)*k!=a[j])t++; if (t<min) { min=t;mini=i; } } printf("%d\n",min); for (i=1;i<=n;i++) if (a[i]!=mini+(i-1)*k) { if (a[i]>mini+(i-1)*k)printf("- %d %d\n",i,a[i]-mini-(i-1)*k); if (a[i]<mini+(i-1)*k)printf("+ %d %d\n",i,-a[i]+mini+(i-1)*k); } return 0; }
C
Problem description
给出一个只有正数和0的边长为n的方形矩阵a,若有a^k全是正数,则打yes,否则打出no(k为任意自然数)
Data Limit：n <=2000 Time Limit: 1s
Solution
由题意,可以把原矩阵看做邻接矩阵,然后只要判断原图是否联通就好了.
Code
#include<cstdio> int n,i,j,k=0; int a[6000][6000]; bool bo[10000000],ans; void dfs1(int x) { if (bo[x])return; bo[x]=true; for (int i=1;i<=n;i++)if (a[i][x]>0)dfs1(i); return; } void dfs2(int x) { if (bo[x])return; bo[x]=true; for (int i=1;i<=n;i++)if (a[x][i]>0)dfs2(i); return; } int main() { scanf("%d",&n); for (i=1;i<=n;i++) for (j=1;j<=n;j++) { scanf("%d",&a[i][j]); //if (a[i][j]>0&&i!=j) //add(i,j); } /*for (i=1;i<=n;i++) { h=0,t=0; for (j=1;j<=n;j++)bo[j]=false; dfs(i); for (j=1;j<=n;j++)if (!bo[j]) { puts("NO"); return 0; } }*/ ans=true; dfs1(1); for (i=1;i<=n;i++) { ans&=bo[i]; bo[i]=false; } dfs2(1); for (i=1;i<=n;i++) { ans&=bo[i]; bo[i]=false; } if (ans)puts("YES");else puts("NO"); return 0; }
E
Problem description
给出一个数列,和m个坏指数,进行几次修改,使所有数的f()值之和最大. 修改:取一个数r,将1到r的数全除以其最大公约数. f():f(1)=0,其他的将原数分解质因数后好质数数-坏质数数=其f值.
Data Limit：1 ≤ n, m ≤ 5000 Time Limit: 1s
Solution
从后向前扫,如果在某处修改后对答案有利,就修改,否则不改.
Code
#include<cstdio> int n,m,i,j; long long a[10000],b[10000],g[10000],top=0,p[1000000],minii,ans=0,mini,ff[20000000],hhh; bool bo[10000000],booo; long long gcd(long long a,long long b) { if (b>a)return gcd(b,a); if (b==0)return a;else return gcd(b,a%b); } long long f(long long x) { int ttt=x; if (x==1)return 0; if (x<10000000&&ff[x]!=0)return ff[x]; long long ans=0,i,j; for (i=1;i<=m;i++) if (x>=b[i]) while (x%b[i]==0) { ans--; x=x/b[i]; if (x<10000000&&ff[x]!=0)return ff[x]+ans; } for (i=1;i<=top;i++) if (x>=p[i]) while (x%p[i]==0) { ans++; x=x/p[i]; if (x<10000000&&ff[x]!=0)return ff[x]+ans; } if (x!=1)ans++; if (ttt<10000000)ff[ttt]=ans; return ans; } int main() { scanf("%d%d",&n,&m); for (i=1;i<=n;i++)scanf("%d",&i[a]); for (i=1;i<=m;i++)scanf("%d",&b[i]);g[1]=a[1]; for (i=2;i<=n;i++)g[i]=gcd(g[i-1],a[i]); for (i=2;i<=100000;i++) if (!bo[i]) { bo[i]=true; for (j=2;j<=100000;j++) if (i*j>100000)break;else bo[i*j]=true; booo=false; for (j=1;j<=m;j++) { if (i==b[j])booo=true; } if (!booo) { top++; p[top]=i; } } int div=1; for (i=n;i>=1;i--) { int tmp=f(g[i]/div); if(tmp<0) div = g[i]; a[i] /= div; } for (i=1;i<=n;i++) ans+=f(a[i]); printf("%d",ans); return 0; }
F
Problem description
给出n和p,构造一个无向图,它有2n+p条边,n个点,无重边自环,且每个有k个点的字图中最多有2k+p个点.
Data Limit：5 ≤ n ≤ 24; p ≥ 0 Time Limit: 1s
Solution
比赛时面向样例编程,然后过了...... 那怎么证明这是对的呢? 暂时不知道...
Code
#include<cstdio> int n,i,j,p,a,b,T; int main() { scanf("%d",&T); while (T--) { scanf("%d%d",&n,&p);a=1;b=2; for (i=1;i<=2*n+p;i++) { printf("%d %d\n",a,b); if (b<n)b++;else { a++;b=a+1; } } } return 0; }
