poj水题若干道

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poj1033 Hangover

题目链接：http://poj.org/problem?id=1003

///2014.7.15///poj1003//Accepted    740K    0MS G++ 385B    2014-07-15 19:58:59//题目大意：已知c=1/2+1/3+1/4+....1/(n+1).现给出一个值m，求n的值使得c刚好超过m。#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){    double num;    double sum;    int n;    while( cin>>num && num!=0 ){        n = 2;        sum = 0;        while( sum < num ){            sum += 1.0/n;            n++;        }        printf("%d card(s)\n",n-2 );    }    return 0;}

poj1004 Financial Management

题目链接：http://poj.org/problem?id=1004

///2014.7.15///poj1004//题目大意：给出12个数，求这12个数的平均数。#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){    double sum = 0;    double num;    int n = 12;    while( n-- ){        cin>>num;        sum += num;    }    cout<<"$"<<sum/12<<endl;    return 0;}

poj1005 I Think I Need a Houseboat

题目链接：http://poj.org/problem?id=1005

///2014.7.15///poj1005//题目大意：已知一个圆心为(0,0),//半径随时间增长的位于X轴上方的半圆，//初始面积为0，每年的面积增加50，给出一个坐标//求该坐标在第几年被该半圆覆盖。#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    // freopen("in","r",stdin);    // freopen("out","w",stdout);    int cas = 0;    int t;    scanf("%d",&t);    double x,y,Area;    while( t-- ){        cas++;        cin>>x>>y;        Area = (x*x+y*y) * 3.141592654 / 2;        int n = 0;        while( n*50 < Area )            n++;        cout<<"Property "<<cas<<": This property will begin eroding in year "<<n<<"."<<endl;    }    cout<<"END OF OUTPUT."<<endl;    return 0;}

poj1207 The 3n + 1 problem

题目链接：http://poj.org/problem?id=1207

///2014.7.15///poj1207// 大致题意：// 根据给定的算法，可以计算一个整数的循环数// 现在给定一个区间，计算这个区间的所有数的循环数，把最大的循环数输出// PS:输出的是整数A的循环数，而不是输出整数A#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int cnum(int i){    int re = 1;    while( i!=1 ){        if( i%2 )            i = i*3+1;        else            i = i/2;        re++;    }    return re;}int main(){    // freopen("in","r",stdin);    // freopen("out","w",stdout);    int a,b;    int maxcnum;    while( cin>>a>>b ){        int x=a<b?a:b;        int y=a>b?a:b;        maxcnum = 0;        for(int i=x ; i<=y ; i++){            int temp = cnum(i);            if( temp > maxcnum )                maxcnum = temp;        }        cout<<a<<' '<<b<<' '<<maxcnum<<endl;    }    return 0;}

poj3299 Humidex

题目链接：http://poj.org/problem?id=3299

#include <iostream>#include <math.h>#include <string>#include <iomanip>using namespace std;int main(){    char alpha;    double t,d,h;    int i;    while( true ){        t=d=h=200;        for(i=0;i<2;i++)        {            cin>>alpha;            if(alpha=='E')                return 0;            else if(alpha=='T')                cin>>t;            else if(alpha=='D')                cin>>d;            else if(alpha=='H')                cin>>h;        }        if( h == 200 )            h=t+0.5555*(6.11*exp(5417.7530*(1/273.16-1/(d+273.16)))-10);        else if(t==200)            t=h-0.5555*(6.11*exp(5417.7530*(1/273.16-1/(d+273.16)))-10);        else if(d==200)            d=1/((1/273.16)-((log((((h-t)/0.5555)+10.0)/6.11))/5417.7530))-273.16;        cout<<setprecision(1)<<fixed<<"T "<<t<<" D "<<d<<" H "<<h<<endl;    }    return 0;}

poj2159 Ancient Cipher

题目链接：http://poj.org/problem?id=2159

思路：只要原字符串跟加密之后的字符串能够一一对应，并且每个字符对应的字符不同即可。

///2014.7.16///poj2159//Accepted  704K    63MS    G++ 797B    2014-07-16 09:45:12#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    // freopen("in","r",stdin);    // freopen("out","w",stdout);    string x,y;    char xnum[26],ynum[26];    for(int i=0 ; i<26 ; i++)        xnum[i] = ynum[i] = 0;    cin>>x>>y;    for(int i=0 ; i<x.length() ; i++)        xnum[ x[i]-'A' ] ++;    for(int i=0 ; i<y.length() ; i++)        ynum[ y[i]-'A' ] ++;    sort(xnum,xnum+26);    sort(ynum,ynum+26);    bool can = true;    for(int i=0 ; i<26 ; i++){        if( xnum[i] != ynum[i] ){            can = false;            break;        }    }    if( can )        cout<<"YES"<<endl;    else        cout<<"NO"<<endl;    return 0;}

poj3094 Quicksum 很水很水

题目链接：http://poj.org/problem?id=3094

///2014.7.17///poj3094//Accepted    380K    0MS G++ 454B    2014-07-17 16:02:43#include <cstdio>char temp;int sum;int main(){    while( ~scanf("%c",&temp) && temp!='#' ){        int i=1;        sum = 0;        while( temp==' ' || ( temp<='Z' && temp>='A' ) ){            if( temp<='Z' && temp>='A' )                sum += (i++) * ( temp-'A'+1 );            else                i++;            scanf("%c",&temp);        }        printf("%d\n",sum );    }    return 0;}

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