51Nod1116 K进制下的大数(思维)

题意：一个大数是k进制并且是(k - 1)的倍数，求出最小的k;

思路：当为k进制的时候，因为k % (k - 1) == 1,k ^ k % (k - 1) == 1,……，所以这个大数每位的权取模之后全是一，所以要判断k进制数是否是(k - 1)的倍数的话直接将各个数相加，判断一下是否是(k - 1)的倍数就行。

COde:

#include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;typedef long long ll;const int maxn = 1000000 + 2;#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x)char s[maxn];int main(){    while( ~ scanf("%s",s))    {        int len = strlen(s);        int sum = 0;        int maxs = 1;        for(int i = 0; i < len; i ++)        if(s[i] >= '0' && s[i] <= '9')            sum += s[i] - '0',maxs = max(maxs,s[i] - '0');        else sum += s[i] - 'A' + 10,maxs = max(maxs,s[i] -'A' + 10 );        int ans = -1;        for(int i = max(2,maxs + 1); i <= 36; i ++)        {            if(sum % (i - 1) == 0)            {                ans = i;                break;            }        }        if(ans == -1)            puts("No Solution");        else cout << ans << endl;    }    return 0;}