Problem: Given two integers ‘n’ and ‘sum’, find count of all n digit numbers with sum of digits as ‘sum’. Leading 0’s are not counted as digits. Print your output % 10^9+7. #include <stdio.h> #define MOD 1000000007 unsigned long long dp[101][1001]; unsigned long long Count(int n,int sum) { int i; if(n==0) { if(sum==0) return 1; return
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