title: woj1002-Genesis
date: 2020-03-05
categories: acm
tags: [acm,woj]
输入输出考虑一下。easy
#include <iostream>
#include <string>
#include<cctype>
using namespace std;// 考虑换行.最后 .和字母在一起不考虑。cin空格截断，刚好int main (){
int first=1,cnt=0;
string s;
while(cin>>s){ //EOF 考虑换行,换行后输出上一组的cnt.first=1表示第一行
if(isdigit(s[0])){
if (first)
{ cout<<s<<" "; first=0;}
else{
cout<<cnt<<endl; cnt=0;cout<<s<<" "; }
}
else if(isalpha(s[0])) cnt++; //s[0] >= 'a' && s[0] <= 'z') || (s[0] >= 'A' && s[0] <= 'Z')
}
cout<<cnt<<endl; //EOF
return 0;
}
title: woj1003-birthofnoah
date: 2020-03-04
categories: acm
tags: [acm,woj,数据结构]
简单题。用到map，pair，typedef，iterater。
好久没做，生疏了好多。
description：略
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>using namespace std;typedef pair<int,int> info; //ancestor,age
map<string,info> family = {{"Adam",make_pair(1,930)},{"Seth",make_pair(2,912)},
{"Enosh",make_pair(3,905)},{"Kenan",make_pair(4,910)},
{"Mahalalel",make_pair(5,895)},{"Jared",make_pair(6,962)},{"Enoch",make_pair(7,365)},{"Methuselah",make_pair(8,969)},
{"Lamech",make_pair(9,777)},{"Noah",make_pair(10,-1)},{"Shem",make_pair(11,-1)},{"Ham",make_pair(11,-1)},{"Japheth",make_pair(11,-1)}};/*好像struct更简单。但是查起来麻烦
struct people {
string name[20];
int ancestor;
int old;
}
*/void initial(){
//复习一下
//info p;
//map<string,info> person;
//p=make_pair(1,930); //p.first,p.second
//person.insert(make_pair("Adam",p)); return;
}int main(){
/*
string line;
while(getline(cin,line,'#'))
{cout <<line;
fflush(stdin);
}
*/
//initial();
//char name1[15],name2[15];
string name1,name2;
map<string,info>::iterator iter1,iter2;
while(cin>>name1>>name2) //cin空格截断.getline可以一行
{
// if(family.count(name1)>0) 必定存在，不判断
iter1=family.find(name1);
iter2=family.find(name2);
if(iter1->second.first==-1||iter2->second.first==-1) //注意map用-> pair用.
cout<<"No enough information"<<endl;
else if(iter1->second.first<iter2->second.first)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;if(iter1->second.second==-1||iter2->second.second==-1)
printf("No enough information\n");
else if(iter1->second.second>iter2->second.second)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}title: woj1004-noah’s ark
date: 2020-03-05
categories: acm
tags: [acm,woj]
简单题。
注意浮点数的处理，用eps 1e-9.注意除法变乘法 a/b=6__a=6*b
description：略
// meters =100 cm 1 inch= 2.54cm feet=0.3048 m cubits 45.72cm
// meters/centimeters/inches/cubits/feet
#include<iostream>
#include <cmath>
const double eps=1e-9;using namespace std;//l==w 则spin。否则， Its length to width ratio of exactly six to one provided excellent stability on the high seas.
// fbs abs 求绝对值 cmathdouble trans(double l,string unit){
if (unit[2]=='t')
l*=100;
else if(unit[2]=='c')
l*=2.54;
else if(unit[2]=='b')
l*=45.72;
else if(unit[2]=='e')
l*=30.48;
return l;
}int main(){
string l,w,h;
int flag=0;
double ll,ww,hh;
while(cin>>ll){
cin>>l>>ww>>w>>hh>>h;
/*
if(!flag)
flag=1;
else
cout<<endl;
*/
ll=trans(ll,l);
ww=trans(ww,w);
if(fabs(ll-ww)<eps)
cout<<"Spin"<<endl;
else if(fabs(ll-6.0*ww)<eps) // 考虑 /0。之前先判断==0，没考虑全。
cout<<"Excellent"<<endl;
else
cout<<"Neither"<<endl;
cout<<endl; }
return 0;
}
