给定的范围内[L,R],任务是找到我* countDigits(ⅰ)2对于所有的i的总和∈[L,R],其中countDigits(i)是数字的计数在我。
也就是说,找到:
L * countDigits(L)2 + (L + 1) * countDigits(L + 1)2 + ….. + R * countDigits(R)2.
例子:
Input: L = 8, R = 11
Output: 101
8 * 12 + 9 * 12 + 10 * 22 + 11 * 22 = 8 + 9 + 40 + 44 = 101
Input: L = 98, R = 102
Output: 2
98 * 22 + 99 * 22 + 100 * 32 + 101 * 32 + 102 * 32 = 3515
方法:我们将段[L,R]分成几段相同位数的数字。
[1 – 9], [10 – 99], [100 – 999], [1000 – 9999], [10000 – 99999], [100000 – 999999], [10000000 – 99999999] 等等。
当L和R的长度相同时,所需的总和将为countDigits(L) 2 * (L + R) * (R – L + 1) / 2
证明:
Let [L, R] = [10, 14] where L and R are of the same length i.e. 2.
Therefore, the sum for the segment [L, R] will be 10 * 22 + 11 * 22 + 12 * 22 + 13 * 22 + 14 * 22.
Take 22 common, 22 * (10 + 11 + 12 + 13 + 14) = totalDigits2 * (Sum of AP)
Sum of AP = (no of terms / 2) * (first term + last term) i.e. (R – L + 1) * (L + R) / 2.
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
#define MOD 1000000007
// Function to return the required sum
int rangeSum(int l, int r)
{
int a = 1, b = 9, res = 0;
for (int i = 1; i <= 10; i++) {
int L = max(l, a);
int R = min(r, b);
// If range is valid
if (L <= R) {
// Sum of AP
int sum = (L + R) * (R - L + 1) / 2;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10;
b = b * 10 + 9;
}
return res;
}
// Driver code
int main()
{
int l = 98, r = 102;
cout << rangeSum(l, r);
return 0;
}
Java
// Java implementation of the approach
class GFG {
static final int MOD = 1000000007;
// Function to return the required sum
static int rangeSum(int l, int r)
{
int a = 1, b = 9, res = 0;
for (int i = 1; i <= 10; i++) {
int L = Math.max(l, a);
int R = Math.min(r, b);
// If range is valid
if (L <= R) {
// Sum of AP
int sum = (L + R) * (R - L + 1) / 2;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10;
b = b * 10 + 9;
}
return res;
}
// Driver code
public static void main(String args[])
{
int l = 98, r = 102;
System.out.print(rangeSum(l, r));
}
}
Python3
# Python3 implementation of the approach
MOD = 1000000007;
# Function to return the required sum
def rangeSum(l, r) :
a = 1; b = 9; res = 0;
for i in range(1, 11) :
L = max(l, a);
R = min(r, b);
# If range is valid
if (L <= R) :
# Sum of AP
sum = (L + R) * (R - L + 1) // 2;
res += (i * i) * (sum % MOD);
res %= MOD;
a *= 10;
b = b * 10 + 9;
return res;
# Driver code
if __name__ == "__main__" :
l = 98 ; r = 102;
print(rangeSum(l, r));
# This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
class GFG {
const int MOD = 1000000007;
// Function to return the required sum
static int rangeSum(int l, int r)
{
int a = 1, b = 9, res = 0;
for (int i = 1; i <= 10; i++) {
int L = Math.Max(l, a);
int R = Math.Min(r, b);
// If range is valid
if (L <= R) {
// Sum of AP
int sum = (L + R) * (R - L + 1) / 2;
res += (i * i) * (sum % MOD);
res %= MOD;
}
a = a * 10;
b = b * 10 + 9;
}
return res;
}
// Driver code
public static void Main()
{
int l = 98, r = 102;
Console.WriteLine(rangeSum(l, r));
}
}
PHP
Javascript
3515
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