题目链接：http://poj.org/problem?id=1152
An Easy Problem!
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 7708 | Accepted: 1847 |
Description
Have you heard the fact "The base of every normal number system is 10" ? Of course, I am not talking about number systems like Stern Brockot Number System. This problem has nothing to do with this fact but may have some similarity.
You will be given an N based integer number R and you are given the guaranty that R is divisible by (N-1). You will have to print the smallest possible value for N. The range for N is 2 <= N <= 62 and the digit symbols for 62 based number is (0..9 and A..Z and a..z). Similarly, the digit symbols for 61 based number system is (0..9 and A..Z and a..y) and so on.
You will be given an N based integer number R and you are given the guaranty that R is divisible by (N-1). You will have to print the smallest possible value for N. The range for N is 2 <= N <= 62 and the digit symbols for 62 based number is (0..9 and A..Z and a..z). Similarly, the digit symbols for 61 based number system is (0..9 and A..Z and a..y) and so on.
Input
Each line in the input will contain an integer (as defined in mathematics) number of any integer base (2..62). You will have to determine what is the smallest possible base of that number for the given conditions. No invalid number will be given as input. The largest size of the input file will be 32KB.
Output
If number with such condition is not possible output the line "such number is impossible!" For each line of input there will be only a single line of output. The output will always be in decimal number system.
Sample Input
3 5 A
Sample Output
4 6 11
题意： 正如题目所说，一个简单的问题！输入一个2-62进制的数字，即假设这个数字是n进制，则这个数x%（n-1）==0。
好吧，既然如此简单，就不废话了。。。。
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; char a[100000]; int main() { int i,s,len,max; bool d; while(scanf("%s",a)!=EOF) { len=strlen(a); s=0; max=0; //max记录所能代表的最小进制数(比如数字里面有B,则这个进制至少也是12吧。。。) for(i=0;i<len;i++) //接下来的过程,不要怀疑.经过验证了的 { //(如果这个n进制的数能取余n-1等于0,则这个数的各个位数相加也要能取余n-1等于0,这样问题变简单咯)[这个很容易证明的,不妨试试] if (a[i]>='0'&&a[i]<='9') { if (a[i]-'0'>max) max=a[i]-'0'; s=s+a[i]-'0'; } if (a[i]>='a'&&a[i]<='z') { if (a[i]-'a'+36>max) max=a[i]-'a'+36; s=s+a[i]-'a'+36; } if (a[i]>='A'&&a[i]<='Z') { if (a[i]-'A'+10>max) max=a[i]-'A'+10; s=s+a[i]-'A'+10; } } d=false; for(i=2;i<=62;i++) if (s%(i-1)==0&&max<i) {d=true;break;} if (d) printf("%d\n",i); else printf("such number is impossible!\n"); } return 0; }
