51nod 1116 K进制下的大数 暴力/数论

题目链接:

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1116

题意:

题解:

这里写图片描述
这里写图片描述

代码:

暴力：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 1e5+10;
17 
18 string str;
19 int len;
20 
21 bool check(int k){
22     ll sum=0;
23     for(int i=0; i<len; i++){
24         if(str[i]>='A') sum =  sum*k + (str[i]-'A'+10);
25         else sum = sum*k + (str[i]-'0');
26         sum %= (k-1);
27     }
28     return sum%(k-1) == 0;
29 }
30 
31 int main(){
32     cin >> str;
33     int mx=-1;
34     len=str.size();
35     for(int i=0; i<len; i++)
36         mx = max(mx,str[i]-'A'+10);
37 
38     int k,f=0;
39     for(k=mx+1; k<=36; k++){
40         if(check(k)) {f=1; break;}
41     }
42 
43     if(f) cout << k << endl;
44     else cout << "No Solution" << endl;
45 
46 
47     return 0;
48 }

 

数学：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 1e5+10;
17 
18 string str;
19 ll sum;
20 
21 int main(){
22     cin >> str;
23     int mx=-1,len = str.size();
24     for(int i=0; i<len; i++){
25         if(str[i]>='A') mx=max(mx,str[i]-'A'+10),sum += str[i]-'A'+10;
26         else mx=max(mx,str[i]-'0'),sum += str[i]-'0';
27     }
28 
29     int f = 0,ans;
30     for(int i=mx; i<36; i++){
31         if(sum % i == 0){
32             f = 1;
33             ans = i+1;
34             break;
35         }
36     }
37 
38     if(f) cout << ans << endl;
39     else cout << "No Solution" << endl;
40 
41     return 0;
42 }

 

posted @ 2017-03-22 20:37  _yxg123  阅读(141)  评论(0)  编辑  收藏  举报