Bestcoder #54 div2

A：A problem of sorting

题目即题意。

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<map>#include<vector>#include<set>#include<ctime>#define LL long long#define db double#define EPS 1e-15#define pa pair<int,int>using namespace std;const int N = 120;  const int inf = 2147483647;  const int mod = 2009;  struct person  {      char name[N];      int year;  }s[N];  bool cmp(person x,person y)  {      return x.year>y.year;  }  int main()  {      int t,n,i,j,l;      scanf("%d",&t);      while(t--)      {          scanf("%d",&n);          getchar();          for(i=0;i<n;i++)          {              s[i].year=0;              gets(s[i].name);              l=strlen(s[i].name);              for(j=l-4;j<l;j++)                  s[i].year=s[i].year*10+s[i].name[j]-'0';              s[i].name[l-5]='\0';          }          sort(s,s+n,cmp);          for(i=0;i<n;i++)              printf("%s\n",s[i].name);      }      return 0;  }  

B：The Factor

题目大意：求一串数乘积的因子中的最小合数；

解题思路：就每个数都分解分解，然后保留下最小的两个质因子。

<pre name="code" class="cpp">#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<map>#include<vector>#include<set>#include<ctime>#define LL long long#define db double#define EPS 1e-15#define pa pair<int,int>

int min1,min2;void fun(__int64 s){ for(__int64 i=2;i*i<=s;i++) while(s%i==0) { if(min1>i||min2>i) { if(min2>min1) min2=i; else min1=i; } s/=i; } if(s>1) if(min1>s||min2>s) { if(min2>min1) min2=s; else min1=s; }}int main(){ int t,n,i; __int64 s; scanf("%d",&t); while(t--) { scanf("%d",&n);min1=min2=inf; for(i=0;i<n;i++) { scanf("%I64d",&s); fun(s); } if(min1==inf||min2==inf) puts("-1"); else printf("%I64d\n",(__int64)min1*min2); } return 0;}

C：Geometric Progression

题目大意：判断一个数列是否是等比数列。

解题思路：检验对所有$1<i<nA[i-1]\ast A[i+1]=A[i]\ast A[i]$ 是否都成立。可以套一个大数模板。

注意……0不能是等比数列的首项……高中课本知识。

/*+,-,*,/,% 可直接使用.CIN读入bignum数据类型*/#include <iostream>#include <string.h>#include<stdio.h>#include<iostream>using namespace std;#define DIGIT    4#define DEPTH    10000#define MAX     100typedef int bignum_t[MAX+1];int read(bignum_t a,istream& is=cin){    char buf[MAX*DIGIT+1],ch;    int i,j;    memset((void*)a,0,sizeof(bignum_t));    if (!(is>>buf))    return 0;    for (a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)        ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch;    for (a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0');    for (i=1;i<=a[0];i++)        for (a[i]=0,j=0;j<DIGIT;j++)            a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0';    for (;!a[a[0]]&&a[0]>1;a[0]--);    return 1;}void write(const bignum_t a,ostream& os=cout){    int i,j;    for (os<<a[i=a[0]],i--;i;i--)        for (j=DEPTH/10;j;j/=10)            os<<a[i]/j%10;}int comp(const bignum_t a,const bignum_t b){    int i;    if (a[0]!=b[0])        return a[0]-b[0];    for (i=a[0];i;i--)        if (a[i]!=b[i])            return a[i]-b[i];    return 0;}int comp(const bignum_t a,const int b){    int c[12]={1};    for (c[1]=b;c[c[0]]>=DEPTH;c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++);    return comp(a,c);}int comp(const bignum_t a,const int c,const int d,const bignum_t b){    int i,t=0,O=-DEPTH*2;    if (b[0]-a[0]<d&&c)        return 1;    for (i=b[0];i>d;i--){        t=t*DEPTH+a[i-d]*c-b[i];        if (t>0) return 1;        if (t<O) return 0;    }    for (i=d;i;i--){        t=t*DEPTH-b[i];        if (t>0) return 1;        if (t<O) return 0;    }    return t>0;}void add(bignum_t a,const bignum_t b){    int i;    for (i=1;i<=b[0];i++)        if ((a[i]+=b[i])>=DEPTH)            a[i]-=DEPTH,a[i+1]++;    if (b[0]>=a[0])        a[0]=b[0];    else        for (;a[i]>=DEPTH&&i<a[0];a[i]-=DEPTH,i++,a[i]++);    a[0]+=(a[a[0]+1]>0);}void add(bignum_t a,const int b){    int i=1;    for (a[1]+=b;a[i]>=DEPTH&&i<a[0];a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++);    for (;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);}void sub(bignum_t a,const bignum_t b){    int i;    for (i=1;i<=b[0];i++)        if ((a[i]-=b[i])<0)            a[i+1]--,a[i]+=DEPTH;    for (;a[i]<0;a[i]+=DEPTH,i++,a[i]--);    for (;!a[a[0]]&&a[0]>1;a[0]--);}void sub(bignum_t a,const int b){    int i=1;    for (a[1]-=b;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);    for (;!a[a[0]]&&a[0]>1;a[0]--);}void sub(bignum_t a,const bignum_t b,const int c,const int d){    int i,O=b[0]+d;    for (i=1+d;i<=O;i++)        if ((a[i]-=b[i-d]*c)<0)            a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH;    for (;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++);    for (;!a[a[0]]&&a[0]>1;a[0]--);}void mul(bignum_t c,const bignum_t a,const bignum_t b){    int i,j;    memset((void*)c,0,sizeof(bignum_t));    for (c[0]=a[0]+b[0]-1,i=1;i<=a[0];i++)        for (j=1;j<=b[0];j++)            if ((c[i+j-1]+=a[i]*b[j])>=DEPTH)                c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH;    for (c[0]+=(c[c[0]+1]>0);!c[c[0]]&&c[0]>1;c[0]--);}void mul(bignum_t a,const int b){    int i;    for (a[1]*=b,i=2;i<=a[0];i++){        a[i]*=b;        if (a[i-1]>=DEPTH)            a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH;    }    for (;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++);    for (;!a[a[0]]&&a[0]>1;a[0]--);}void mul(bignum_t b,const bignum_t a,const int c,const int d){    int i;    memset((void*)b,0,sizeof(bignum_t));    for (b[0]=a[0]+d,i=d+1;i<=b[0];i++)        if ((b[i]+=a[i-d]*c)>=DEPTH)            b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH;    for (;b[b[0]+1];b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH);    for (;!b[b[0]]&&b[0]>1;b[0]--);}void div(bignum_t c,bignum_t a,const bignum_t b){    int h,l,m,i;    memset((void*)c,0,sizeof(bignum_t));    c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1;    for (i=c[0];i;sub(a,b,c[i]=m,i-1),i--)        for (h=DEPTH-1,l=0,m=(h+l+1)>>1;h>l;m=(h+l+1)>>1)            if (comp(b,m,i-1,a)) h=m-1;            else l=m;    for (;!c[c[0]]&&c[0]>1;c[0]--);    c[0]=c[0]>1?c[0]:1;}void div(bignum_t a,const int b,int& c){    int i;    for (c=0,i=a[0];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--);    for (;!a[a[0]]&&a[0]>1;a[0]--);}void sqrt(bignum_t b,bignum_t a){    int h,l,m,i;    memset((void*)b,0,sizeof(bignum_t));    for (i=b[0]=(a[0]+1)>>1;i;sub(a,b,m,i-1),b[i]+=m,i--)        for (h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1)            if (comp(b,m,i-1,a)) h=m-1;            else l=m;    for (;!b[b[0]]&&b[0]>1;b[0]--);    for (i=1;i<=b[0];b[i++]>>=1);}int length(const bignum_t a){    int t,ret;    for (ret=(a[0]-1)*DIGIT,t=a[a[0]];t;t/=10,ret++);    return ret>0?ret:1;}int digit(const bignum_t a,const int b){    int i,ret;    for (ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT;i;ret/=10,i--);    return ret%10;}int zeronum(const bignum_t a){    int ret,t;    for (ret=0;!a[ret+1];ret++);    for (t=a[ret+1],ret*=DIGIT;!(t%10);t/=10,ret++);    return ret;}void comp(int* a,const int l,const int h,const int d){    int i,j,t;    for (i=l;i<=h;i++)        for (t=i,j=2;t>1;j++)            while (!(t%j))                a[j]+=d,t/=j;}void convert(int* a,const int h,bignum_t b){    int i,j,t=1;    memset(b,0,sizeof(bignum_t));    for (b[0]=b[1]=1,i=2;i<=h;i++)        if (a[i])            for (j=a[i];j;t*=i,j--)                if (t*i>DEPTH)                    mul(b,t),t=1;    mul(b,t);}void combination(bignum_t a,int m,int n){    int* t=new int[m+1];    memset((void*)t,0,sizeof(int)*(m+1));    comp(t,n+1,m,1);    comp(t,2,m-n,-1);    convert(t,m,a);    delete []t;}void permutation(bignum_t a,int m,int n){    int i,t=1;    memset(a,0,sizeof(bignum_t));    a[0]=a[1]=1;    for (i=m-n+1;i<=m;t*=i++)        if (t*i>DEPTH)            mul(a,t),t=1;    mul(a,t);}#define SGN(x) ((x)>0?1:((x)<0?-1:0))#define ABS(x) ((x)>0?(x):-(x))int read(bignum_t a,int &sgn,istream& is=cin){    char str[MAX*DIGIT+2],ch,*buf;    int i,j;    memset((void*)a,0,sizeof(bignum_t));    if (!(is>>str)) return 0;    buf=str,sgn=1;    if (*buf=='-') sgn=-1,buf++;    for (a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--)        ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch;    for (a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0');    for (i=1;i<=a[0];i++)        for (a[i]=0,j=0;j<DIGIT;j++)            a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0';    for (;!a[a[0]]&&a[0]>1;a[0]--);    if (a[0]==1&&!a[1]) sgn=0;    return 1;}struct bignum{    bignum_t num;    int sgn;public:inline bignum(){memset(num,0,sizeof(bignum_t));num[0]=1;sgn=0;}//inline int operator!(){return num[0]==1&&!num[1];}inline bignum& operator=(const bignum& a){memcpy(num,a.num,sizeof(bignum_t));sgn=a.sgn;return *this;}inline bignum& operator=(const int a){memset(num,0,sizeof(bignum_t));num[0]=1;sgn=SGN(a);add(num,sgn*a);return *this;};inline bignum& operator+=(const bignum& a){if(sgn==a.sgn)add(num,a.num);else if(sgn&&a.sgn){int ret=comp(num,a.num);if(ret>0)sub(num,a.num);else if(ret<0){bignum_t t;    memcpy(t,num,sizeof(bignum_t));memcpy(num,a.num,sizeof(bignum_t));sub(num,t);sgn=a.sgn;}else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0;}else if(!sgn)memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn;return *this;}inline bignum& operator+=(const int a){if(sgn*a>0)add(num,ABS(a));else if(sgn&&a){int ret=comp(num,ABS(a));if(ret>0)sub(num,ABS(a));else if(ret<0){bignum_t t;    memcpy(t,num,sizeof(bignum_t));memset(num,0,sizeof(bignum_t));num[0]=1;add(num,ABS(a));sgn=-sgn;sub(num,t);}else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0;}else if(!sgn)sgn=SGN(a),add(num,ABS(a));return *this;}inline bignum operator+(const bignum& a){bignum ret;memcpy(ret.num,num,sizeof(bignum_t));ret.sgn=sgn;ret+=a;return ret;}inline bignum operator+(const int a){bignum ret;memcpy(ret.num,num,sizeof(bignum_t));ret.sgn=sgn;ret+=a;return ret;}inline bignum& operator-=(const bignum& a){if(sgn*a.sgn<0)add(num,a.num);else if(sgn&&a.sgn){int ret=comp(num,a.num);if(ret>0)sub(num,a.num);else if(ret<0){bignum_t t;    memcpy(t,num,sizeof(bignum_t));memcpy(num,a.num,sizeof(bignum_t));sub(num,t);sgn=-sgn;}else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0;}else if(!sgn)add(num,a.num),sgn=-a.sgn;return *this;}inline bignum& operator-=(const int a){if(sgn*a<0)add(num,ABS(a));else if(sgn&&a){int ret=comp(num,ABS(a));if(ret>0)sub(num,ABS(a));else if(ret<0){bignum_t t;    memcpy(t,num,sizeof(bignum_t));memset(num,0,sizeof(bignum_t));num[0]=1;add(num,ABS(a));sub(num,t);sgn=-sgn;}else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0;}else if(!sgn)sgn=-SGN(a),add(num,ABS(a));return *this;}inline bignum operator-(const bignum& a){bignum ret;memcpy(ret.num,num,sizeof(bignum_t));ret.sgn=sgn;ret-=a;return ret;}inline bignum operator-(const int a){bignum ret;memcpy(ret.num,num,sizeof(bignum_t));ret.sgn=sgn;ret-=a;return ret;}inline bignum& operator*=(const bignum& a){bignum_t t;mul(t,num,a.num);memcpy(num,t,sizeof(bignum_t));sgn*=a.sgn;return *this;}inline bignum& operator*=(const int a){mul(num,ABS(a));sgn*=SGN(a);return *this;}inline bignum operator*(const bignum& a){bignum ret;mul(ret.num,num,a.num);ret.sgn=sgn*a.sgn;return ret;}inline bignum operator*(const int a){bignum ret;memcpy(ret.num,num,sizeof(bignum_t));mul(ret.num,ABS(a));ret.sgn=sgn*SGN(a);return ret;}inline bignum& operator/=(const bignum& a){bignum_t t;div(t,num,a.num);memcpy(num,t,sizeof(bignum_t));sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn;return *this;}inline bignum& operator/=(const int a){int t;div(num,ABS(a),t);sgn=(num[0]==1&&!num[1])?0:sgn*SGN(a);return *this;}inline bignum operator/(const bignum& a){bignum ret;bignum_t t;memcpy(t,num,sizeof(bignum_t));div(ret.num,t,a.num);ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn;return ret;}inline bignum operator/(const int a){bignum ret;int t;memcpy(ret.num,num,sizeof(bignum_t));div(ret.num,ABS(a),t);ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a);return ret;}inline bignum& operator%=(const bignum& a){bignum_t t;div(t,num,a.num);if (num[0]==1&&!num[1])sgn=0;return *this;}inline int operator%=(const int a){int t;div(num,ABS(a),t);memset(num,0,sizeof(bignum_t));num[0]=1;add(num,t);return t;}inline bignum operator%(const bignum& a){bignum ret;bignum_t t;memcpy(ret.num,num,sizeof(bignum_t));div(t,ret.num,a.num);ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn;return ret;}inline int operator%(const int a){bignum ret;int t;memcpy(ret.num,num,sizeof(bignum_t));div(ret.num,ABS(a),t);memset(ret.num,0,sizeof(bignum_t));ret.num[0]=1;add(ret.num,t);return t;}inline bignum& operator++(){*this+=1;return *this;}inline bignum& operator--(){*this-=1;return *this;};inline int operator>(const bignum& a){return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0);}inline int operator>(const int a){return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0);}inline int operator>=(const bignum& a){return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0);}inline int operator>=(const int a){return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0);}inline int operator<(const bignum& a){return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0);}inline int operator<(const int a){return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0);}inline int operator<=(const bignum& a){return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0);}inline int operator<=(const int a){return sgn<0?(a<0?comp(num,-a)>=0:1):(sgn>0?(a>0?comp(num,a)<=0:0):a>=0);}inline int operator==(const bignum& a){return (sgn==a.sgn)?!comp(num,a.num):0;}inline int operator==(const int a){return (sgn*a>=0)?!comp(num,ABS(a)):0;}inline int operator!=(const bignum& a){return (sgn==a.sgn)?comp(num,a.num):1;}inline int operator!=(const int a){return (sgn*a>=0)?comp(num,ABS(a)):1;}inline int operator[](const int a){return digit(num,a);}friend inline istream& operator>>(istream& is,bignum& a){read(a.num,a.sgn,is);return is;}friend inline ostream& operator<<(ostream& os,const bignum& a){if(a.sgn<0)os<<'-';write(a.num,os);return os;}friend inline bignum sqrt(const bignum& a){bignum ret;bignum_t t;memcpy(t,a.num,sizeof(bignum_t));sqrt(ret.num,t);ret.sgn=ret.num[0]!=1||ret.num[1];return ret;}friend inline bignum sqrt(const bignum& a,bignum& b){bignum ret;memcpy(b.num,a.num,sizeof(bignum_t));sqrt(ret.num,b.num);ret.sgn=ret.num[0]!=1||ret.num[1];b.sgn=b.num[0]!=1||ret.num[1];return ret;}inline int length(){return ::length(num);}inline int zeronum(){return ::zeronum(num);}inline bignum C(const int m,const int n){combination(num,m,n);sgn=1;return *this;}inline bignum P(const int m,const int n){permutation(num,m,n);sgn=1;return *this;}};#define N 105bignum s[N];int main(){    int t,n,i,k;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=k=0;i<n;i++)        {            cin>>s[i];            if(s[i]==0)                k++;        }        if(k>0&&k<n)        {            puts("No");            continue;        }        if(n<3)        {            puts("Yes");            continue;        }        for(i=1;i<n-1;i++)            if(s[i-1]*s[i+1]!=s[i]*s[i])                break;        if(i<n-1)            puts("No");        else            puts("Yes");    }    return 0;}

D：Reflect

题目大意：圆上一点发出光线，有多少种情况能返回原点。

解题思路：发射射线与其在该点相切的线的夹角记作sita,则每条边所对应的圆心角为2*sita,反射n次有n+1条边，则2*sita*(n+1)=2*k*pi,k为整数；

sita=pi*k/(n+1);sita的种类数与k与(n+1)互质的情况数相同，所以转化为求k与n+1互质的情况数；

暴力gcd即可。

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>#include<map>#include<vector>#include<set>#include<ctime>#define LL long long#define db double#define EPS 1e-15#define inf 1e16#define pa pair<int,int>using namespace std;int t,n,m;int gcd(int a,int b){  if(b==0) return a;  return gcd(b,a%b);}int main(){  int i,j,k,ans;  scanf("%d",&t);  while(t--){    scanf("%d",&n);    ans=0;    for(i=1;i<=n+1;i++){        if(gcd(i,n+1)==1){            ans++;        }    }    printf("%d\n",ans);  }  return 0;}