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

 

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 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 }