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.
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#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 0; } if(dp[n][sum]!=-1) return dp[n][sum]; unsigned long long res=0; for(i=0;i<=9;i++) { if(sum-i>=0) res = (res+Count(n-1,sum-i))%MOD; } return dp[n][sum]=res; } unsigned long long countDigit(int n,int sum) { int i; if(n==0) { if(sum==0) return 1; return 0; } unsigned long long res=0; for(i=1;i<=9;i++) { if(sum-i>=0) res = (res+Count(n-1,sum-i))%MOD; } return res; } int main() { //code int t,n,sum,i,j; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&sum); for(i=0;i<=n;i++) { for(j=0;j<=sum;j++) dp[i][j]=-1; } int d = countDigit(n,sum); if(d==0) d=-1; printf("%d\n",d); } return 0; } |