构造水题 Codeforces Round #206 (Div. 2) A. Vasya and Digital Root

 1 /*
 2     构造水题：对于0的多个位数的NO，对于位数太大的在后面补0，在9×k的范围内的平均的原则
 3 */
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstring>
 7 #include <cmath>
 8 using namespace std;
 9 
10 const int MAXN = 1e3 + 10;
11 const int INF = 0x3f3f3f3f;
12 int a[MAXN];
13 
14 int main(void)        //Codeforces Round #206 (Div. 2) A. Vasya and Digital Root
15 {
16     //freopen ("A.in", "r", stdin);
17     int k, d;
18     while (scanf ("%d%d", &k, &d) == 2)
19     {
20         if (d == 0 && k > 1)    {puts ("No solution");    continue;}
21         if (k >= d)
22         {
23             for (int i=1; i<=d; ++i)    putchar (‘1‘);
24             for (int i=1; i<=k-d; ++i)    putchar (‘0‘);
25             puts ("");
26         }
27         else
28         {
29             if (9 * k < d)    {puts ("No solution");    continue;}
30             for (int i=1; i<=k; ++i)    a[i] = d / k;
31             int res = d % k;
32             for (int i=1; i<=res; ++i)    a[i]++;
33             for (int i=1; i<=k; ++i)    printf ("%d", a[i]);
34             puts ("");
35         }
36 
37     }
38 
39     return 0;
40 }