woj1002-Genesis woj1003-birthofnoah woj1004-noah's ark

---

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 cmath

double 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;
}