PAT 1010. Radix (25)

此题高能，甲级里面难度算比较大的了！（其实也不难，关键没指明数据范围）

思路：

  最后的答案不止是≤36，也可能特别大（超过int， 达到 long long int）。不能用顺序查找，需用二分查找。

这题有一个特点：当待处理的字符串只有一位时，答案有多个；当长度大于一位时，只有一个正确答案（或者没有答案）。

注意

当radix很大的时候，可能会导致计算过程超出long long，需要判断如果是负数，就说明这个radix太大了，应该向更小的查找。

AC代码

#define LOCAL
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define Pii pair<int, int>
#define MAX_RADIX 1223372036854775807
typedef long long LL;
const int MAXN = 10+5;
const LL MAX_INTEGER = 9223372036854775807;

int get_dig(char ch) {
int dig;
if(ch >= '0' && ch <= '9')
dig = ch - '0';
else
dig = ch - 'a' + 10;
return dig;
}

LL get_num(char s[], LL r, LL goal = MAX_INTEGER) {
int n = strlen(s);
LL res = 0;
LL power = r;
for(int i = n-1; i >= 0; i--) {
int dig = get_dig(s[i]);
if(i == n-1) {
res = dig;
} else {
if(r < 0) return -1;    //too big
res += dig * power;
power *= r;
}
if(res < 0 || res > goal) return -1;    //too big
}
return res;
}

LL binary_search_radix(char s[], LL goal) {
LL low = 0, high = MAX_RADIX;
int n = strlen(s);
//找到可能的最小radix
for(int i = 0; i < n; i++) {
low = max(low, (LL)get_dig(s[i]));
}
low = low + 1;

while(low <= high) {
LL mid = low + (high - low) / 2;
LL res = get_num(s, mid, goal);
if(res == goal) return mid;
else if(res == -1) high = mid - 1;
else low = mid + 1;
}
return -1;
}

int main() {
#ifdef LOCAL
freopen("../DATA/data.in", "r", stdin);
freopen("../DATA/data.out", "w", stdout);
#endif
char a[MAXN], b[MAXN], *c;
int tag, radix;
while(scanf("%s%s%d%d", a, b, &tag, &radix) == 4) {
LL goal;
LL r;
if(tag == 1) {
goal = get_num(a, radix);
c = b;
} else {
goal = get_num(b, radix);
c = a;
}

//特殊处理只有一位的情况
if(strlen(c) == 1) {
if(get_dig(c[0]) == goal) r = goal + 1;
else r = -1;
}
else {
r = binary_search_radix(c, goal);
}

if(r == -1) printf("Impossible\n");
else printf("%lld\n", r);
}
return 0;
}

如有不当之处欢迎指出！