**Sam and substrings**

### Problem Statement :

Samantha and Sam are playing a numbers game. Given a number as a string, no leading zeros, determine the sum of all integer values of substrings of the string. Given an integer as a string, sum all of its substrings cast as integers. As the number may become large, return the value modulo 10^9 +7. Example n = '42' Here n is a string that has 3 integer substrings: 4, 2, and 42. Their sum is 48, and 48 modulo{10^9 +7)=48. Function Description Complete the substrings function in the editor below. substrings has the following parameter(s): string n: the string representation of an integer Returns int: the sum of the integer values of all substrings in n, modulo 10^9 + 7 Input Format A single line containing an integer as a string, without leading zeros. Constraints 1 <= ncastasaninteger <= 2 x 10^5

### Solution :

` ````
Solution in C :
In C++ :
#include <iostream>
#include <string.h>
using namespace std;
#define MOD 1000000007
int main() {
char num[200005];
gets(num);
int len = strlen(num);
long long factor = 1, ans = 0;
for (int i = len-1; i >= 0; --i) {
long long tmp = (num[i]-'0') * (i+1) * factor % MOD;
ans += tmp;
ans %= MOD;
factor = (factor * 10 + 1) % MOD;
}
printf("%d\n", ans);
return 0;
}
In Java :
import java.util.Scanner;
public class Solution {
static long mod = 1000000007;
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
char s[] = cin.next().toCharArray();
long ten[] = new long[s.length + 2];
ten[0] = 1;
for (int i=1; i<ten.length; i++) {
ten[i] = ten[i - 1] * 10 + 1;
ten[i] %= mod;
}
long ans = 0;
for (int i=0; i<s.length; i++) {
ans += ((s[i] - '0') * ten[s.length - i - 1]) % mod * (i + 1) % mod;
ans %= mod;
}
System.out.println(ans);
cin.close();
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXLEN 200001
#define MODULUS 1000000007LL
int main() {
long long total, multiplier, partialsum;
char c;
total = partialsum = 0;
multiplier = 1;
while (1) {
c=getchar();
if (c<'0' || c>'9') break;
c -= '0';
partialsum += multiplier * c;
total = (total * 10 + partialsum) % MODULUS;
multiplier++;
}
printf("%lld\n",total);
return 0;
}
In Python3 :
import operator
MOD = 1000000007
s = input()
n = len(s)
a = []
p10 = 0
for i in range(n, 0, -1):
p10 = (p10 * 10 + 1) % MOD
a.append(p10 * i % MOD)
b = [ord(c) - ord('0') for c in s]
print(sum(map(operator.mul, reversed(a), b)) % MOD)
```

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