一些做的过的零散的题目(poj)（二）:

1003:

比较简单:之前是在ubuntu下写的程序，但是后来编码被我搞乱了。一直ac不了。就换成vs2010了

#include <stdio.h>int main(){double c;while((scanf("%lf",&c)!=EOF) && c!=0){double sum = 0.0;int i=1;while(1){sum += 1.0/(i+1);if(sum >= c)break;i++;}printf("%d card(s)\n",i);}return 0;}

1006：

生物钟：运用中国剩余定理。计算过程：提供的三个周期分别为：23,28,33。

假设a,b,c使：(33*28*a) % 23 = 1(23*33*b) % 28 =1(28*23*c)%33 = 1

容易求出：a=6 b=19 c=2 

33*28*6 = 5544 23*33*19 = 14421  28*23*2 = 1288

所以ans = (5544*p + 14421*e + 1288*i - d) % (23*28*33)

#include <stdio.h>int main(){int p,e,i,d;int time=1;while(scanf("%d%d%d%d",&p,&e,&i,&d)){if(p==-1 && e==-1 && i==-1 && d==-1)break;int lcm=21252;int n=(5544*p+14421*e+1288*i-d+21252)%21252;if(n==0)n=21252;printf("Case %d: the next triple peak occurs in %d days.\n",time++,n);}return 0;}

1007:

DNA排序：首先获取输入并求出每个序列的秩，并用short数组存放。然后按照冒泡法（稳定）排序：按照秩降序排序并输出。

#include <iostream>#include <string.h>using namespace std;int get_NUM(char* str){int num = 0;for(int j=1;j<strlen(str);j++)for(int k=0;k<j;k++){if(str[k]>str[j])num++;}return num;}int main(){// ifstream ifs;// ifs.open("test.txt");int n,m;cin>>n>>m;char str[102][52];unsigned short snum[101];for(int i=0;i<m;i++){cin>>str[i];str[i][101]=get_NUM(str[i]);snum[i]=get_NUM(str[i]);}unsigned short stemp;char temp[101];for(int i=0;i<m;i++)for(int j=m-1;j>i;j--){if(snum[j] < snum[j-1]){strcpy(temp,str[j-1]);strcpy(str[j-1],str[j]);strcpy(str[j],temp);stemp = snum[j-1];snum[j-1] = snum[j]; snum[j] = stemp;}}for(int i=0;i<m;i++){//cout<<(int)(str)[i][n]<<endl;str[i][n]='\0';str[i][n+1] = '\0';cout<<str[i]<<endl;}return 0;}

1008：

日历题：求出以0年0月1日为起点计算出输入日期距离它的天数，然后通过求余和除法的到具体日期。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;char Haab[20][8] ={"","pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax","zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};char Tzolkin[21][10] = {"","imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};int main(){int num;cin>>num;int m;for(int i=0;i<num;i++){int day=0,year=0,month=0;char months[10]={};char days[5]={};cin>>days>>months>>year;for(int i=0;i<strlen(days)-1;i++)day = day*10 + days[i]-'0';if(i==0)cout<<num<<endl;//cout<<day<<months<<year<<endl;//cout<<day<<endl<<months<<endl<<year<<endl;for(int j=1;j<=19;j++)if(strcmp(months,Haab[j])==0){month = j; break;}m = year*365 + day + (month-1) * 20;//以0年0月1日为起点，计算距离天数。// 0 - 364天//cout<<m<<endl;year = m / 260;//259 m %= 260;month = m%13 + 1;day = m%20 + 1;cout<<month<<" "<<Tzolkin[day]<<" "<<year<<endl;}}

2080:

    也是一个日期题但是计算方法不太一样：将输入的日期N 按照是否是闰年 减掉365或者366。 然后按照月份的31 29 30天来减掉月份，然后剩下的就是日期数目。

   
#include <iostream>#include <stdio.h>using namespace std;int leap(int year){if(year%4!=0 || (year%100==0&&year%400!=0))return 0;return 1;}int years[2]={365,366};int months[2][12]={31,28,31,30,31,30,31,31,30,31,30,31,  31,29,31,30,31,30,31,31,30,31,30,31};char weeks[7][10] = {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};int main(){int n;while(cin>>n && n!=-1){int month,year,week;week = n%7;for(year=2000;n>=years[leap(year)];year++)n -= years[leap(year)];for(month=0;n>=months[leap(year)][month];month++)n -= months[leap(year)][month]; printf("%d-%02d-%02d %s\n",year,month+1,n+1,weeks[(week+5)%7]);     }return 0;}