这次的百度之星，不得不吐槽下系统的判题数据，被坑了不知多少次。
第一题：
大意：求一段区间的累乘。用线段树即可。
坑点：如果询问范围超出边界，输出上一次的结果。
/*
Problem :
Status :
By wf,
*/
#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
typedef long long ll;
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=9973;
const int maxn = 1e6+5;
char str[1000000+5];
int sum[maxn<<2];
int index;
int len;
void PushUP(int rt)
{
sum[rt] =( sum[rt<<1] * sum[rt<<1|1] )%mod;
}
void build(int l,int r,int rt)
{
if(l==r)
{
//printf("%c ",str[index]);
sum[rt] = str[index++]-28;
}
if (l == r) return ;
int m = (l + r) >> 1;
build(lson);
build(rson);
PushUP(rt);
}
int query(int L,int R,int l,int r,int rt)
{
if (L <= l && r <= R)
{
return sum[rt];
}
int m = (l + r) >> 1;
int ret = 1;
if (L <= m) ret = (ret * query(L , R , lson) )%mod ;
if (R > m) ret = (ret* query(L , R , rson) ) %mod;
//printf("ret==%d
",ret);
return ret;
}
int main()
{
//freopen("in_a.txt","r",stdin);
int n,a,b;
int ans = 0;
while (~scanf("%d",&n))
{
index=0;
scanf("%s",&str);
len = strlen(str);
build(1 , len , 1);
int a,b;
for(int i=0;i<n;++i)
{
scanf("%d %d",&a,&b);
if(a>=1 && a<=len && 1<=b && b<=len) ans = query(a,b,1,len,1);
printf("%d
", query(a,b,1,len,1) );
}
}
return 0;
}
第二题：
多写几组数据就能发现：是求大数版的斐波那切数列。
坑点：如果询问的是0，则输出空行。
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
/* * 完全大数模板 * 输出cin>>a * 输出a.print(); * 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。 */
#define MAXN 9999
#define MAXSIZE 1010
#define DLEN 4
class BigNum
{
private:
int a[500]; //可以控制大数的位数
int len;
public:
BigNum()
{
len=1; //构造函数
memset(a,0,sizeof(a));
}
BigNum(const int); //将一个int类型的变量转化成大数
BigNum(const char*); //将一个字符串类型的变量转化为大数
BigNum(const BigNum &); //拷贝构造函数
BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算
friend istream& operator>>(istream&,BigNum&); //重载输入运算符
friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符
BigNum operator+(const BigNum &)const; //重载加法运算符，两个大数之间的相加运算
BigNum operator-(const BigNum &)const; //重载减法运算符，两个大数之间的相减运算
BigNum operator*(const BigNum &)const; //重载乘法运算符，两个大数之间的相乘运算
BigNum operator/(const int &)const; //重载除法运算符，大数对一个整数进行相除 运算
BigNum operator^(const int &)const; //大数的n次方运算
int operator%(const int &)const; //大数对一个int类型的变量进行取模运算
bool operator>(const BigNum &T)const; //大数和另一个大数的大小比较
bool operator>(const int &t)const; //大数和一个int类型的变量的大小比较
void print(); //输出大数
};
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
int c,d=b;
len=0;
memset(a,0,sizeof(a));
while(d>MAXN)
{
c=d-(d/(MAXN+1))*(MAXN+1);
d=d/(MAXN+1);
a[len++]=c;
}
a[len++]=d;
}
BigNum::BigNum(const char *s) //将一个字符串类型的变量转化为大数
{
int t,k,index,L,i;
memset(a,0,sizeof(a));
L=strlen(s);
len=L/DLEN;
if(L%DLEN)len++;
index=0;
for(i=L-1; i>=0; i-=DLEN)
{
t=0;
k=i-DLEN+1;
if(k<0)k=0;
for(int j=k; j<=i; j++) t=t*10+s[j]-'0';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum &T):len(T.len) //拷贝构造函数
{
int i;
memset(a,0,sizeof(a));
for(i=0; i<len; i++) a[i]=T.a[i];
}
BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符，大数之间赋值运算
{
int i;
len=n.len;
memset(a,0,sizeof(a));
for(i=0; i<len; i++) a[i]=n.a[i];
return *this;
}
istream& operator>>(istream &in,BigNum &b)
{
char ch[MAXSIZE*4];
int i=-1;
in>>ch;
int L=strlen(ch);
int count=0,sum=0;
for(i=L-1; i>=0;)
{
sum=0;
int t=1;
for(int j=0; j<4&&i>=0; j++,i--,t*=10)
{
sum+=(ch[i]-'0')*t;
}
b.a[count]=sum;
count++;
}
b.len=count++;
return in;
}
ostream& operator<<(ostream& out,BigNum& b) //重载输出运算符
{
int i;
cout<<b.a[b.len-1];
for(i=b.len-2; i>=0; i--)
{
printf("%04d",b.a[i]);
}
return out;
}
BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算
{
BigNum t(*this);
int i,big;
big=T.len>len?T.len:len;
for(i=0; i<big; i++)
{
t.a[i]+=T.a[i];
if(t.a[i]>MAXN)
{
t.a[i+1]++;
t.a[i]-=MAXN+1;
}
}
if(t.a[big]!=0) t.len=big+1;
else t.len=big;
return t;
}
BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=0;
}
else
{
t1=T;
t2=*this;
flag=1;
}
big=t1.len;
for(i=0; i<big; i++)
{
if(t1.a[i]<t2.a[i])
{
j=i+1;
while(t1.a[j]==0) j++;
t1.a[j--]--;
while(j>i) t1.a[j--]+=MAXN;
t1.a[i]+=MAXN+1-t2.a[i];
}
else t1.a[i]-=t2.a[i];
}
t1.len=big;
while(t1.a[len-1]==0 && t1.len>1)
{
t1.len--;
big--;
}
if(flag) t1.a[big-1]=0-t1.a[big-1];
return t1;
}
BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i=0; i<len; i++)
{
up=0;
for(j=0; j<T.len; j++)
{
temp=a[i]*T.a[j]+ret.a[i+j]+up;
if(temp>MAXN)
{
temp1=temp-temp/(MAXN+1)*(MAXN+1);
up=temp/(MAXN+1);
ret.a[i+j]=temp1;
}
else
{
up=0;
ret.a[i+j]=temp;
}
}
if(up!=0) ret.a[i+j]=up;
}
ret.len=i+j;
while(ret.a[ret.len-1]==0 && ret.len>1)ret.len--;
return ret;
}
BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算
{
BigNum ret;
int i,down=0;
for(i=len-1; i>=0; i--)
{
ret.a[i]=(a[i]+down*(MAXN+1))/b;
down=a[i]+down*(MAXN+1)-ret.a[i]*b;
}
ret.len=len;
while(ret.a[ret.len-1]==0 && ret.len>1) ret.len--;
return ret;
}
int BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模
{
int i,d=0;
for(i=len-1; i>=0; i--) d=((d*(MAXN+1))%b+a[i])%b;
return d;
}
BigNum BigNum::operator^(const int &n)const //大数的n次方运算
{
BigNum t,ret(1);
int i;
if(n<0)exit(-1);
if(n==0)return 1;
if(n==1)return *this;
int m=n;
while(m>1)
{
t=*this;
for(i=1; (i<<1)<=m; i<<=1) t=t*t;
m-=i;
ret=ret*t;
if(m==1)ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum &T)const //大数和另一个大数的大小比较
{
int ln;
if(len>T.len)return true;
else if(len==T.len)
{
ln=len-1;
while(a[ln]==T.a[ln]&&ln>=0) ln--;
if(ln>=0 && a[ln]>T.a[ln]) return true;
else return false;
}
else return false;
}
bool BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this>b;
}
void BigNum::print() //输出大数
{
int i;
printf("%d",a[len-1]);
for(i=len-2; i>=0; i--) printf("%04d",a[i]);
printf("
");
}
BigNum f[250];//卡特兰数
int main()
{
//freopen("in_b.txt","r",stdin);
f[1]="1";
f[2]="2";
for(int i=3; i<=210; i++) f[i]=f[i-1]+f[i-2];
int n;
while(cin>>n)
{
if(n==0)
{
cout<<endl;
continue;
}
cout<<f[n]<<endl;
}
return 0;
}
第三题：
字典树。
/*
Problem :
Status :
By wf,
*/
#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
typedef long long ll;
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1e5+5;
#define MAX 26
struct Trie
{
Trie *next[MAX];
int v; //根据需要变化
Trie()
{
v=0;
for(int i=0; i<MAX; ++i)
{
next[i]=NULL;
}
}
};
Trie *root;
void Insert(char *str)
{
//printf("insert:%s
",str);
int len = strlen(str);
Trie *p = root, *q;
for(int i=0; i<len; ++i)
{
int id = str[i]-'a';
if(p->next[id]==NULL)
p->next[id]=new Trie();
p=p->next[id];
p->v++;
}
}
int Search(char *str)
{
//printf("search:%s
",str);
int len = strlen(str);
Trie *p = root;
for(int i=0; i<len; ++i)
{
int id = str[i]-'a';
p = p->next[id];
if(p == NULL) //若为空集，表示不存以此为前缀的串
return 0;
//printf("%c v:%d
",str[i],p->v);
}
return p->v;
}
void Delete(char * str)
{
//printf("delete:%s
",str);
int len = strlen(str);
//取删除单词的个数
int num = Search(str);
Trie *p = root;
/*for(int i=0; i<len; ++i)
{
int id = str[i]-'a';
p = p->next[id];
if(p == NULL) //若为空集，表示不存以此为前缀的串
break ;
//printf("->%c ",str[i]);
num = min(num,p->v) ;
}*/
//printf("num==%d
",num);
//这次再删
p = root;
for(int i=0; i<len; ++i)
{
int id = str[i]-'a';
p = p->next[id];
if(p == NULL) //若为空集，表示不存以此为前缀的串
return ;
//printf("->%c ",str[i]);
p->v -= num;
if(p->v ==0)
{
for(int i=0; i<26; ++i)
{
p->next[i]=NULL;
}
p=NULL;
return;
}
}
return ;
}
int main()
{
//freopen("in_c.txt","r",stdin);
int n;
scanf("%d",&n);
char cmd[20],str[5000];
int len;
root = new Trie();
while(n--)
{
//cin>>cmd>>str;
scanf("%s %s",&cmd,&str);
switch( cmd[0] )
{
case 'i':
Insert(str);
break;
case 'd':
Delete(str);
break;
case 's':
if( Search(str) )
{
cout<<"Yes"<<endl;
}
else cout<<"No"<<endl;
break;
}
}
return 0;
}
第四题：
水题，string排序+map。
/*
Problem :
Status :
By wf,
*/
#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
typedef long long ll;
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1e5+5;
map<string,int>mp;
int main()
{
//freopen("in_d.txt","r",stdin);
int n;
string str;
cin>>n;
while(n--)
{
cin>>str;
sort(str.begin(),str.end() );
cout<<mp[str]<<endl;
mp[str]++;
}
return 0;
}
第五题：
大意：给你一些语句，如果从第i-1行的语句和第i行的语句不能找到满足条件的变量值，输出第i-1的编号；否则输出unique。
不难，处理好字符串就行了。
/*
Problem :
Status :
By wf,
*/
#include "algorithm"
#include "iostream"
#include "cstring"
#include "cstdio"
#include "string"
#include "stack"
#include "cmath"
#include "queue"
#include "set"
#include "map"
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
typedef long long ll;
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1e5+5;
struct node
{
string name;
int l,r;
} var[50005];
struct node2
{
int num;
int varid[50];
bool flag;
} p[1005];
map<string,int>mp;
bool iskong(int l1,int r1,int l2,int r2)
{
if( l2>r1 || l1>r2 )return 1;
return 0;
}
int main()
{
//freopen("in_e.txt","r",stdin);
string str;
int n;
cin>>n;
getchar();
int id = 1;
for(int ii=0; ii<n; ii++)
{
getline(cin,str);
//cout<<str<<endl;
int i=0,j,len = str.length();
p[ii].num=0;
p[ii].flag=false;
while(i<len)
{
while(i<len && str[i]==' ' )i++;
if( isalpha(str[i]) )
{
//获取变量名
j=i+1;
while(j<len && isalpha(str[j]) )++j;
var[id].name = str.substr(i,j-i);
i=j;
//获取运算符
int ops=0;
while(i<len && str[i]==' ' )i++;
if( str[i]=='<' && str[i+1]!='=' )
{
ops=1;
i++;
}
else if( str[i]=='>' && str[i+1]!='=' )
{
ops=2;
i++;
}
else if( str[i]=='<' && str[i+1]=='=' )
{
ops=3;
i+=2;
}
else if( str[i]=='>' && str[i+1]=='=' )
{
ops=4;
i+=2;
}
else if( str[i]=='=' && str[i+1]=='=' )
{
ops=5;
i+=2;
}
//获取数字
int num=0;
while(i<len && str[i]==' ')i++;
if( isdigit(str[i]) )
{
num = str[i]-'0';
j=i+1;
while(j<len && isdigit(str[j]) )
{
num = num*10 + str[j] - '0';
j++;
}
i=j;
}
//处理
switch(ops)
{
case 1: //<
var[id].l = -10000;
var[id].r = num-1;
break;
case 2: //>
var[id].l = num+1;
var[id].r = 10000;
break;
case 3: //<=
var[id].l = -10000;
var[id].r = num;
break;
case 4: //>=
var[id].l = num;
var[id].r = 10000;
break;
case 5: //==
var[id].l = num;
var[id].r = num;
break;
}
//判断这个条件是否为空
//printf("dangqian num:%d
",p[ii].num);
for(int t=0; t<p[ii].num; t++)
{
int tid = p[ii].varid[t];
//cout<<"name1:"<<var[id].name<<" name2:"<<var[tid].name<<endl;
if( var[id].name == var[tid].name && iskong(var[tid].l,var[tid].r,var[id].l,var[id].r) )
{
p[ii].flag=1;
//printf("debug
");
}else
{
var[id].l = max( var[id].l , var[tid].l );
var[tid].l = var[id].l;
var[id].r = min( var[id].r , var[tid].r );
var[tid].r = var[id].r;
}
}
//处理完毕
//cout<<name<<" ops=="<<ops<<" num=="<<num<<" l=="<<var[id].l<<" r=="<<var[id].r<<" flag=="<<p[ii].flag<<endl;
p[ii].varid[ p[ii].num++]=id;
id++;
while(i<len && str[i]==' ')i++;
if( i<len && str[i]==',' )
{
i++;
}
}
else i++;
}
if( p[ii].flag )
{
cout<<"unique"<<endl;
continue;
}
bool isUnique=true;
bool first = 1;
for(int jj=0; jj<ii; ++jj)
{
bool isThisOk =1;
if( p[jj].flag )continue;
else
{
for(int k1 = 0; k1<p[ii].num; k1++)
{
for(int k2 = 0; k2<p[jj].num; k2++)
{
if(!isThisOk )break;
int tid1 = p[ii].varid[k1];
int tid2 = p[jj].varid[k2];
//printf("[%d,%d] [%d,%d]
",var[tid1].l,var[tid1].r,var[tid2].l,var[tid2].r);
//cout<<"name1:"<<var[tid1].name<<" name2:"<<var[tid2].name<<endl;
if( var[tid1].name == var[tid2].name && iskong(var[tid1].l,var[tid1].r,var[tid2].l,var[tid2].r) )
{
isThisOk=0;
//printf("ok
");
break;
}
}
}
}
if( isThisOk )
{
isUnique=false;
if(first)first=0;
else cout<<" ";
cout<<jj+1;
}
}
if( isUnique )cout<<"unique"<<endl;
else cout<<endl;
//cout<<endl;
}
return 0;
return 0;
}