POJ 2545+2591+2247+1338简单水题

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【题意简述】：就是有这样的一个序列，就拿当p1 = 2,p2 = 3, p3 = 5,来举例，由这三个数为基准组成的序列是:

2,3,4,5,6,8,9,10,12……现在给你这个序列数组的下标，让你求得这个数组中，这个下标里的数是多少。

【分析】：2,3,4,5,6,8,9,10,12……这个序列式由2,3,5这三个数生成的，具体如何生成，就是：

详见代码：

这里以POJ2545为例：

//弄清其中的逻辑关系，可以从最简单的2,3,5试着做起！ #include<iostream>#include<algorithm>using namespace std;int main(){long long p1,p2,p3;int n;long long H[100005];int a1,a2,a3;while(cin>>p1>>p2>>p3>>n){H[1] = 1;a1=a2=a3=1;for(int i = 2;i<=n+1;i++){H[i] = min(p1*H[a1],min(p2*H[a2],p3*H[a3]));if(H[i]==p1*H[a1])  a1++;if(H[i]==p2*H[a2])  a2++;if(H[i]==p3*H[a3])  a3++;}cout<<H[n+1]<<endl;}return 0;}

另外三道题与这个相差无几，都是这个的变形

只不过2247 注意输出的格式！

//  C++ 代码 题目本身很简单，注意格式输出！！ #include<iostream>#include<algorithm>#include<cstring>using namespace std;int main(){int a1,a2,a3,a4;int n;long long humble[6000];humble[1]=1;a1=1;a2=1;a3=1;a4=1;for(int i=2;i<=5842;i++){humble[i] = min(2*humble[a1],min(3*humble[a2],min(5*humble[a3],7*humble[a4])));if(humble[i]==2*humble[a1])a1++;if(humble[i]==3*humble[a2])a2++;if(humble[i]==5*humble[a3])a3++;if(humble[i]==7*humble[a4])a4++;}//string b;while(1){cin>>n;if(n==0)break;if(n%10==1){if(n%100==11)cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;elsecout<<"The "<<n<<"st"<<" humble number is "<<humble[n]<<"."<<endl;}if(n%10==2){if(n%100==12)cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;elsecout<<"The "<<n<<"nd"<<" humble number is "<<humble[n]<<"."<<endl;}if(n%10==3){if(n%100==13)cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;elsecout<<"The "<<n<<"rd"<<" humble number is "<<humble[n]<<"."<<endl;}if(n%10>3||n%10==0)cout<<"The "<<n<<"th"<<" humble number is "<<humble[n]<<"."<<endl;}}

POJ 2591

// 39392K  141Ms// 打表过得 =_= #include<iostream>#include<algorithm>using namespace std;int S[10000001];int main(){int a,b,n;S[1] = 1;a = b = 1;for(int i = 2;i<=10000000;i++){S[i] = min(2*S[a]+1,3*S[b]+1);if(S[i] == 2*S[a]+1) a++;if(S[i] == 3*S[b]+1) b++;}while(cin>>n){cout<<S[n]<<endl;}return 0;}

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