1010. Radix (25)

参考http://www.cnblogs.com/549294286/p/3571604.html

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 N1 and N2, 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

/**
2017.3.7
Donald
*/

//1010. Radix (25)

/**
题意：input：N1，N2，Tag, Radix
if Tag == 1，find K，if N1(Radix) == N2(K)
若是  ：输出最小的K
若不是：输出Impossible
if Tag == 2, find K, if N2(Radix) == N1(K)
若是  ：输出最小的K
若不是：输出Impossible
*/
/**
思路：
若tag == 2， 交换N1，N2
1.将N1转化为P的进制数
使用按权展开
2.找到K，能将N2转化为N1(Radix)。
找上线，即N1(Radix)
找下线，从0 ~ N( 每个digit > N，N++)

另外测试用例：
1.若N1 = N2  = 1， print 2；
2.若N1 = N2 != 1， print radix;
*/

#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long
#define MAXN 11

char N1[MAXN], N2[MAXN];
int Tag;
LL Radix;

void solve();
LL changeDec();
LL findLowRadix();
LL binarySearch(LL low, LL high, LL top);
int compare(LL radix, LL target);

int main(void){

scanf("%s %s %d %lld", N1, N2, &Tag, &Radix);

if(Tag == 2){
char q[MAXN];
strcpy(q, N1);
strcpy(N1, N2);
strcpy(N2, q);
}
//    for(int i = 0 ; i < strlen(N1); ++i){
//        printf("%c ", N1[i]);
//    }
//    printf("\n");
//    for(int i = 0 ; i < strlen(N2); ++i){
//        printf("%c ", N2[i]);
//    }

solve();

return 0;
}

void solve(){
LL target = changeDec();
LL least  = findLowRadix();
LL most   = (target + 1 > least + 1) ? target + 1 : least + 1;  //eg. 1, so +1;
LL res = binarySearch(least, most, target);
if(res == -1)
printf("Impossible");
else
printf("%lld", res);

}

LL changeDec(){
LL res = 0;
LL cnt = 1;
LL num;
for(LL i = strlen(N1) - 1; i >= 0; --i){
if(N1[i] >= 'a' && N1[i] <= 'z')
num = N1[i] - 'a' + 10;
else if(N1[i] >= '0' && N1[i] <= '9')
num = N1[i] - '0';
res += num * cnt;
cnt *= Radix;
}
return res;
}

LL findLowRadix(){
LL low = 0;
LL num;
for(LL i = strlen(N2) - 1; i >= 0; --i){
if(N2[i] >= 'a' && N2[i] <= 'z')
num = N2[i] - 'a' + 10;
else if(N2[i] >= '0' && N2[i] <= '9')
num = N2[i] - '0';
if(num + 1 > low)
low = num + 1;
}
return low;
}

LL binarySearch(LL low, LL high, LL top){
LL mid = low;
LL tmp;

while(low <= high){
tmp = compare(mid, top);
if(tmp>0){
high = mid - 1;
}else if(tmp < 0 ){
low = mid + 1;
}
else
return mid;

mid = (low + high) / 2;
}
return -1;
}

int compare(LL radix, LL target){
LL cnt = 1;
LL num = 1;
LL sum = 0;
for(LL i = strlen(N2) -1; i >= 0; --i){
if(N2[i] >= 'a' && N2[i] <= 'z')
num = N2[i] - 'a' + 10;
else if(N2[i] >= '0' && N2[i] <= '9')
num = N2[i] - '0';
sum += num * cnt;
cnt *= radix;
if(sum > target)
return 1;
}
if(sum > target)
return 1;
else if(sum < target)
return -1;
else
return 0;
}