题目:click me~
题意:输入4个正整数N1,N2,tag,radix,tag=1表示N1为radix进制数,tag=2表示N2为radix进制数。N1,N2不超过10个数位,每个数位均为0~9或a~z。0~9表示数字0~9,a~z表示数字10~35。
求未知进制的那个数是否存在和另一个数在十进制下相等的进制,若存在,则输出满足条件的进制;否则输出“Impossible”。
解题思路:
步骤一:将已知进制的数放在N1,未知进制的数放在N2,便于后面统一计算。
步骤二:将N1转换为十进制,使用long long 存储(可能有10个数位,36进制,会超过int,但不会超过long long)对一个给定数字串来说,进制越大,那么转换成十进制的结果也越大,因此可以使用二分法。二分N2的进制,将N2从该进制转换成十进制与N1比较:如果大于N1的十进制,说明N2当前进制太大,应该往左子区间继续二分;如果小于,说明N2当前进制太小,应该往右子区间继续二分。二分结束时可以判断解是否存在。
code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef long long LL;
LL map[256];
LL inf = (1LL << 63) - 1;//long long 最大值2^63-1
void init() {
for (char c = '0';c <= '9';c++) {
map[c] = c - '0';
}
for (char c = 'a';c <= 'z';c++) {
map[c] = c - 'a' + 10;
}
}
LL convertnum10(char a[], LL radix, LL t) {
LL ans = 0;
int len = strlen(a);
for (int i = 0;i < len;i++) {
ans = ans * radix + map[a[i]];
if (ans<0 || ans>t)return -1;//溢出或超过N1的十进制
}
return ans;
}
int cmp(char N2[], LL radix, LL t) {
int len = strlen(N2);
LL num = convertnum10(N2, radix, t);
if (num < 0)return 1;
if (t > num)return -1;
else if (t == num)return 0;
else return 1;
}
LL binarysearch(char N2[], LL left, LL right, LL t) {
LL mid;
while (left <= right) {
mid = (left + right) / 2;
int flag = cmp(N2, mid, t);
if (flag == 0)return mid;
else if (flag == -1)left = mid + 1;
else right = mid - 1;
}
return -1;
}
int findld(char N2[]) {
int ans = -1, len = strlen(N2);
for (int i = 0;i < len;i++) {
if (map[N2[i]] > ans) {
ans = map[N2[i]];
}
}
return ans + 1;
}
char N1[20], N2[20], temp[20];
int tag, radix;
int main() {
init();
scanf("%s %s %d %d", N1, N2, &tag, &radix);
if (tag == 2) {
strcpy(temp, N1);
strcpy(N1, N2);
strcpy(N2, temp);
}
LL t = convertnum10(N1, radix, inf);
LL low = findld(N2);
LL high = max(low, t);
LL ans = binarysearch(N2, low, high, t);
if (ans == -1)printf("Impossible\n");
else printf("%lld\n", ans);
return 0;
}