查找数字总和等于值的数字
有一个数字n和一个值。我们必须找到所有n位数字，其中s的所有n位数字之和与给定值相同。此处0不会算作数字。
数字n必须在1到100的范围内，并且值必须在1到500的范围内。
输入输出
Input: This algorithm takes number of digits, and the sum value. Let the digit count is 3. and sum is 15. Output: Display the number of different 3-digit numbers whose sum is 15. The result is 69. (There are 69 different 3-digit numbers whose sum is 15)
算法
计数（数字，总和）
输入： 数字位数，给定值。
输出：计算多少个数字。
Begin
if digit = 0, then
return true when sum = 0
if memTable[digit, sum] is not vacant, then
return memTable[digit, sum]
answer := 0
for i := 0 to 9 do
if sum – i >= 0, then
answer := answer + count(digit – 1, sum - i)
done
return memTable[digit, sum] := answer
EndnumberCount（数字，总和）
输入： 数字位数，给定值。
输出：该类型的数字有多少个数字。
Begin
define memTable and make all space vacant
res := 0
for i := 1 to 9, do
if sum – i >= 0, then
res := res + count(digit – 1, sum - i)
done
return result
End示例
#include<iostream>
#define ROW 101
#define COL 501
using namespace std;
unsigned long long int memTable[ROW][COL];
unsigned long long int count(int digit, int sum) {
if (digit == 0) //for 0 digit number check if the sum is 0 or not
return sum == 0;
if (memTable[digit][sum] != -1) //when subproblem has found the result, return it
return memTable[digit][sum];
unsigned long long int ans = 0; //set the answer to 0 for first time
for (int i=0; i<10; i++) //count for each digit and find numbers starting with it
if (sum-i >= 0)
ans += count(digit-1, sum-i);
return memTable[digit][sum] = ans;
}
unsigned long long int numberCount(int digit, int sum) {
for(int i = 0; i<ROW; i++) //initially all entries of memorization table is -1
for(int j = 0; j<ROW; j++)
memTable[i][j] = -1;
unsigned long long int result = 0;
for (int i = 1; i <= 9; i++) //count for each digit and find numbers starting with it
if (sum-i >= 0)
result += count(digit-1, sum-i);
return result;
}
int main() {
int digit, sum;
cout << "Enter digit count: "; cin >> digit;
cout << "Enter Sum: "; cin >> sum;
cout << "Number of values: " << numberCount(digit, sum);
}输出结果
Enter digit count: 3 Enter Sum: 15 Number of values: 69