题目描述
Today, ZLZX has a mysterious case: Orange lost his down jacket hanging in his dorm room. Under the expectations of everyone, detective Groundhog took his small spoon of the artifact and started the journey to solve the case.
After an in-depth investigation of the northernmost mysterious room on each floor, Groundhog discovered {n}n mysterious numbers. As long as the clues conveyed by these numbers are deciphered, he can reveal the truth of the matter. The deciphering method is: using these numbers to generate two positive integers without leading zeros, and minimizing the product of these two positive integers is the final clue.
Then Groundhog wants to know: What is the smallest product?
As he continued to investigate in the room west of the new building, he gave you the question.
Concise meaning:Given n numbers between 0 and 9, use them to make two positive integers without leading zeros to minimize the product.
输入描述:
The first line of input is a single integer T
T,the number of test cases.
For each set of data：
Each test case begins with a single integer nn, the count of numbers.
The next line are n
n numbers.
输出描述:
For each set of Case, an integer is output, representing the smallest product.
示例1
输入
1
4
1 2 2 1输出
122- 1
示例2
输入
2
5
1 3 2 1 2
3
1 1 0输出
1223
10题解：
本题就是n个（0-9）的数组构成俩个正整数，使得其乘积最小。所以我们很容易想到，使得其中一个数为这些数中最小的一个数，然后使用剩下的数构成一个最小的数。
思路：将其输入在一个数组里面，找出其中最小的数字（不为0）构成第一个数，另一个数就是排序后剩下的数字组合成的最小的数（最高位是排序后数组中第二个不为0的数，然后接上所有零，然后再将剩余的数从小到大拼上就好)
注意的是，这是个大数的乘法！
下面有四个版本的代码
（1）我最初想到的，利用字符串进行储存与运算，但效率不高，而且代码不够精简。
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 9 using namespace std; 10 11 const int N = 1e6; 12 13 inline int read() { 14 int x = 0, f = 1; 15 char ch = getchar(); 16 while(ch<'0'||ch>'9'){ 17 if(ch=='-') 18 f=-1; 19 ch=getchar(); 20 } 21 while(ch>='0'&&ch<='9'){ 22 x = x * 10 + ch - '0'; 23 ch = getchar(); 24 } 25 return x * f; 26 } 27 28 int digit[N]; 29 30 string muil(int a, string b) { 31 string c = ""; 32 int carry_bit = 0; //进位 33 for (int k = b.size() - 1; k >= 0; k--) { 34 int temp = a * (b[k] - '0') + carry_bit; 35 int m = temp % 10; 36 carry_bit = temp / 10; 37 c += (m + '0'); 38 } 39 if (carry_bit != 0) 40 c += (carry_bit + '0'); 41 return c; 42 } 43 44 int main() { 45 int t = read(); 46 while (t--) { 47 fill (digit, digit+N, -1); 48 int n = read(); 49 for (int i = 0; i < n; i++) { 50 digit[i] = read(); 51 } 52 sort(digit, digit+n); 53 54 int zero = 0, one; 55 string anthor = ""; 56 int i = 0; 57 for (; i < n; i++) { 58 if (digit[i] == 0) { 59 zero++; 60 continue; 61 } 62 one = digit[i++]; 63 break; 64 } 65 anthor += (digit[i++] + '0'); 66 while (zero--) { 67 anthor += '0'; 68 } 69 while (i < n) { 70 anthor += (digit[i] + '0'); 71 i++; 72 } 73 74 string res = muil(one, anthor); 75 for (int i = res.size() - 1; i >= 0; i--) { 76 cout << (res[i] - '0'); 77 } 78 cout << "\n"; 79 } 80 }
（2）后来看了别人的博客，发现可以使用一个int型数组存储运算会快一点
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 9 using namespace std; 10 #define ll long long 11 const ll N = 1e5 + 5; 12 13 inline ll read() { 14 ll x = 0, f = 1; 15 char ch = getchar(); 16 while(ch<'0'||ch>'9'){ 17 if(ch=='-') 18 f=-1; 19 ch=getchar(); 20 } 21 while(ch>='0'&&ch<='9'){ 22 x = x * 10 + ch - '0'; 23 ch = getchar(); 24 } 25 return x * f; 26 } 27 28 ll digit[N], res[N]; 29 30 int main() { 31 ll t = read(); 32 while (t--) { 33 fill (digit, digit+N, 0); 34 fill (res, res+N, 0); 35 ll n = read(); 36 for (ll i = 0; i < n; i++) { 37 digit[i] = read(); 38 } 39 sort(digit, digit+n); 40 int pos = 0; 41 while(!digit[pos]) pos++; 42 ll zero = 0, first; 43 zero = pos; 44 first = digit[pos]; 45 46 res[0] = digit[++pos]; 47 ++pos; 48 ll k = 1; //pos记录digit下标, k 是记录res下标 49 50 while (zero--) { 51 res[k++] = 0; 52 } 53 54 for (; pos < n; pos++) { 55 res[k++] = digit[pos]; 56 } 57 58 for (int j = 0; j < n-1; j++) res[j] *= first; 59 for (int j = n-2; j >= 1; j--) { 60 if (res[j] >= 10) { 61 res[j-1] = res[j-1] + res[j] / 10; 62 res[j] = res[j] % 10; 63 } 64 } 65 for (int j = 0; j < k; j++) { 66 cout << res[j]; 67 } 68 cout << "\n"; 69 } 70 } 71 72 /* 73 1 74 7 75 0 0 0 0 1 2 3 76 */
(3) 别人博客中的精简代码
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 9 using namespace std; 10 #define ll long long 11 const ll N = 1e5 + 5; 12 13 inline ll read() { 14 ll x = 0, f = 1; 15 char ch = getchar(); 16 while(ch<'0'||ch>'9'){ 17 if(ch=='-') 18 f=-1; 19 ch=getchar(); 20 } 21 while(ch>='0'&&ch<='9'){ 22 x = x * 10 + ch - '0'; 23 ch = getchar(); 24 } 25 return x * f; 26 } 27 28 ll digit[N]; 29 30 int main() { 31 ll t = read(); 32 while (t--) { 33 fill (digit, digit+N, 0); 34 ll n = read(); 35 36 for (ll i = 0; i < n; i++) { 37 digit[i] = read(); 38 } 39 sort(digit, digit+n); 40 41 int pos = 0; 42 while(!digit[pos]) pos++; 43 44 swap(digit[0],digit[pos]); 45 swap(digit[1],digit[pos+1]); 46 47 for (int j = 1; j < n; j++) digit[j] *= digit[0]; 48 for (int j = n-1; j > 1; j--) { 49 if (digit[j] >= 10) { 50 digit[j-1] = digit[j-1] + digit[j] / 10; 51 digit[j] = digit[j] % 10; 52 } 53 } 54 for (int j = 1; j < n; j++) { 55 cout << digit[j]; 56 } 57 cout << "\n"; 58 } 59 }