2014年携程程序设计大赛 预赛第一场 A,B,C

A. 

题解:

     反面来想...一个数列如果到不了-1..说明这个数列所有数的最大公约数不是1...所以用B^A-到不了的数目=答案..

Program:

#include<iostream>#include<algorithm>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<set>#define ll long long#define oo 1<<29#define MAXN 100000+5using namespace std;ll sum,A,B,P[21]={0,2,3,5,7,11,13,17,19,23};int gcd(int a,int b){       if (a%b==0) return b;       return gcd(b,a%b);}ll POW(ll p,int k){       ll x=1;       while (k--) x*=p;       return x;       }void dfs(ll now,ll data,ll num){       ll m=0,j;        if (B%data!=0) return;       for (j=data;j<=B;j++)          if (j%data==0) m++;       if (num%2) sum+=POW(m,A);             else sum-=POW(m,A);       for (now=now+1;now<=8;now++)              dfs(now,data*P[now],num+1);  }int main(){       ll k,x;        freopen("input.txt","r",stdin);       freopen("output.txt","w",stdout);       scanf("%I64d",&k);       while (k--)       {               scanf("%I64d%I64d",&A,&B);               sum=0;               for (x=1;x<=8;x++) dfs(x,P[x],1);               printf("%I64d\n",POW(B,A)-sum);       }       return 0;}

B.

题解:

     很简单的二位dp..dp[l][r]代表区间[l,r]为一个合法的括号序列所需要加的最少括号..

Program:

#include<iostream>#include<algorithm>#include<string.h>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<set>#define ll long long#define oo 1<<29#define MAXN 100+5using namespace std; int dp[MAXN][MAXN];char s[MAXN];int count(int l,int r){       if (s[l]=='(' && s[r]==')') return 0;       if (s[l]=='[' && s[r]==']') return 0;       return 2;}int main(){        int k,len,LEN,x,l,r,i;       freopen("input.txt","r",stdin);       freopen("output.txt","w",stdout);       scanf("%d",&k);        while (k--)       {               scanf("%s",s+1);               len=strlen(s+1);                memset(dp,0x7f,sizeof(dp));                for (x=1;x<=len;x++) dp[x][x]=1,dp[x+1][x]=0;               for (LEN=1;LEN<len;LEN++)                  for (x=1;x<=len-LEN;x++)                  {                         l=x,r=x+LEN;                         dp[l][r]=dp[l+1][r-1]+count(l,r);                         dp[l][r]=min(dp[l][r],dp[l+1][r]+1);                          dp[l][r]=min(dp[l][r],dp[l][r-1]+1);                         for (i=l;i<r;i++)                            dp[l][r]=min(dp[l][r],dp[l][i]+dp[i+1][r]);                  }                   printf("%d\n",dp[1][len]);       }       return 0;}

C.

题解:

     主要是要算下球面距离公式.然后就是裸的最小生成树了.

Program:

#include<iostream>#include<algorithm>#include<string.h>#include<stdio.h>#include<queue>#include<cmath>#include<stack>#include<map>#include<set>#define ll long long#define oo 1<<29#define MAXN 100+5#define eps 1e-5using namespace std; const double pi=3.14159265358979323846;struct point{       double x,y;}P[MAXN];double D,L,sum,m,A[MAXN][MAXN];bool used[MAXN];double calc(point a,point b){       double ans=(D/2)*acos(sin(a.x)*sin(b.x)+cos(a.x)*cos(b.x)*cos(a.y-b.y));       return ans;}int main(){        int k,C,t,i,j,x;       freopen("input.txt","r",stdin);       freopen("output.txt","w",stdout);       scanf("%d",&k);        while (k--)       {              scanf("%lf%lf",&D,&L);              scanf("%d",&C);              for (i=1;i<=C;i++)              {                     scanf("%lf%lf",&P[i].x,&P[i].y);                     P[i].x=P[i].x/180*pi;                     P[i].y=P[i].y/180*pi;              }              for (i=1;i<=C;i++)                for (j=1;j<=C;j++)                   A[i][j]=calc(P[i],P[j]);              sum=0;              memset(used,0,sizeof(used));              used[1]=true;              for (t=2;t<=C;t++)              {                      m=-1;                      for (i=1;i<=C;i++)                        if (used[i])                          for (j=1;j<=C;j++)                            if (!used[j])                               if (A[i][j]<m || m<0)                                  m=A[i][j],x=j;                      sum+=m;                      used[x]=true;              }              if (L>=sum) puts("Y");                     else puts("N");       }       return 0;}