題干：

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is?yes, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers?N?1???and?N?2??, your task is to find the radix of one number while that of the other is given.

Input Specification:

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:

N1 N2 tag radix

Here?N1?and?N2?each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9,?a-z?} where 0-9 represent the decimal numbers 0-9, and?a-z?represent the decimal numbers 10-35. The last number?radix?is the radix of?N1?if?tag?is 1, or of?N2?if?tag?is 2.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation?N1?=?N2?is true. If the equation is impossible, print?Impossible. If the solution is not unique, output the smallest possible radix.

Sample Input 1:

6 110 1 10

Sample Output 1:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

題目大意：

給定兩個(gè)字符串，每個(gè)都是0~9 ,?a~z 代表0~35這36個(gè)數(shù)字。

給定一個(gè)數(shù)字的radix，你的任務(wù)是找到另一個(gè)數(shù)字的基數(shù)，使得N1=N2。

解題報(bào)告：

我們知道，把十進(jìn)制數(shù)轉(zhuǎn)化成其他進(jìn)制是困難的，但是把其他進(jìn)制轉(zhuǎn)化成十進(jìn)制是較為簡(jiǎn)單的。所以直接鎖定一個(gè)目標(biāo)進(jìn)制之后，轉(zhuǎn)化成十進(jìn)制比較兩個(gè)數(shù)字式是否相同就可以了。剛開(kāi)始想錯(cuò)了，以為就是最多就是35進(jìn)制了，但是其實(shí)則不然，可以是無(wú)窮進(jìn)制，所以不能直接枚舉了。考慮到上界是比較好確定的，并且隨著radix的增大，轉(zhuǎn)化的十進(jìn)制數(shù)是單調(diào)遞增的，考慮二分。

注意二分的上下界限制就好了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; char n1[MAX],n2[MAX],ned[MAX]; int tag,radix; __int128 shi; int go(char c) {if(c >= '0' && c <= '9') return c - '0';else return c - 'a' + 10; } __int128 trans(char s[],ll rdx) {//把s從rdx進(jìn)制轉(zhuǎn)化成10進(jìn)制 __int128 res = 0;int len = strlen(s);for(int i = 0; i<len; i++) {res = res * rdx + go(s[i]);if(res < 0) return (__int128)9e18 * 2;}return res; }int main() {cin>>n1>>n2>>tag>>radix;if(tag == 1) shi = trans(n1,radix),strcpy(ned,n2);else shi = trans(n2,radix),strcpy(ned,n1);int mx = 0,len = strlen(ned);for(int i = 0; i<len; i++) {mx = max(mx,go(ned[i]));}__int128 l = mx+1,r = 9e18,mid;ll ans=-1;while(l<=r) {mid = (l+r)>>1;__int128 tmp = trans(ned,mid); if(tmp < shi) l = mid+1;else r = mid-1;if(tmp == shi) ans = mid;}if(ans != -1) printf("%lld\n",ans);else printf("Impossible\n");return 0 ; }

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