[2016-03-19][UVA][11549][Calculator Conundrum]

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时间:2016-03-19 21:27:43 星期六
题目编号:[2016-03-19][UVA][11549][Calculator Conundrum]
题目大意:给定数k每次取前n位不断平方,求出现的最大值是多少
方法:
- 方法1:模拟一遍过程,直到出现循环
- 方法2:Floyd判断算法,定义两个k,每次k₁走一次,k₂走两次,知道k₁,k₂相同

方法1:STL超级暴力方法

#include <set>
#include <sstream>
#include <string>
#include <cstdio>
using namespace std;
typedef long long LL;
int t,n,k,ans;
int mnxt(){
        stringstream ss;
        ss<<(LL) k * k;
        string str = ss.str();
        if(str.length() > n)    str = str.substr(0,n);
        int num = 0;
        stringstream ss2(str);
        ss2>>num;
        return num;
}
int main(){
        scanf("%d",&t);
        while(t--){
                scanf("%d%d",&n,&k);
                set<int> s;
                ans = k;
                while(!s.count(k)){
                        s.insert(k);
                        if(k > ans)     ans = k;
                        k = mnxt();
                }
                printf("%d\n",ans);
        }
        return 0;
}

方法2:小小优化版

#include <cstdio>
#include <set>
#include<algorithm>
using namespace std;
typedef long long LL;
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))
int t,n,k,ans,buf[20];
int mnxt(){
        if(!k)  return 0;
        LL k2 = (LL)k*k;
        int cnt = 0;
        while (k2){
                buf[cnt++] = k2 % 10;
                k2 /= 10;
        }
        int tmpn = min(n,cnt);     
        FOR(i,0,tmpn){
                k2 = k2 * 10 + buf[--cnt];
        }
        return k2;
}
int main(){
        scanf("%d",&t);
        while(t--){
                scanf("%d%d",&n,&k);
                set<int> s;
                ans = k;
                while(!s.count(k)){
                        s.insert(k);
                        if(k > ans)     ans = k;
                        k = mnxt();
                }
                printf("%d\n",ans);
        }
        return 0;
}

方法3:Floyd判圈算法

#include <cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))
int t,n,ans,buf[20],k1,k2;
int mnxt(int k){
        if(!k)  return 0;
        LL k2 = (LL)k*k;
        int cnt = 0;
        while (k2){
                buf[cnt++] = k2 % 10;
                k2 /= 10;
        }
        int tmpn = min(n,cnt);     
        FOR(i,0,tmpn){
                k2 = k2 * 10 + buf[--cnt];
        }
        return k2;
}
int main(){
        scanf("%d",&t);
        while(t--){
                scanf("%d%d",&n,&k1);               
                k2 = ans = k1;
                do{
                        k1 = mnxt(k1);
                        k2 = mnxt(k2);  if(k2 > ans)    ans = k2;
                        k2 = mnxt(k2);  if(k2 > ans)    ans = k2;
                }
                while(k1 != k2);               
                printf("%d\n",ans);
        }
        return 0;
}