UVA 10198 Counting
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Counting
The Problem
Gustavo knows how to count, but he is now learning how write numbers. As he is a very good student, he already learned 1, 2, 3 and 4. But he didn't realize yet that 4 is different than 1, so he thinks that 4 is another way to write 1. Besides that, he is having fun with a little game he created himself: he make numbers (with those four digits) and sum their values. For instance:
132 = 1 + 3 + 2 = 6112314 = 1 + 1 + 2 + 3 + 1 + 1 = 9 (remember that Gustavo thinks that 4 = 1)After making a lot of numbers in this way, Gustavo now wants to know how much numbers he can create such that their sum is a number n. For instance, for n = 2 he noticed that he can make 5 numbers: 11, 14, 41, 44 and 2 (he knows how to count them up, but he doesn't know how to write five). However, he can't figure it out for n greater than 2. So, he asked you to help him.
The Input
Input will consist on an arbitrary number of sets. Each set will consist on an integer n such that 1 <= n <= 1000. You must read until you reach the end of file.
The Output
For each number read, you must output another number (on a line alone) stating how much numbers Gustavo can make such that the sum of their digits is equal to the given number.
Sample Input
23
Sample Output
513
题意:Gustavo数数时总是把1和4搞混,他认为4只是1的另外一种写法。给出一个整数n,Gustavo想知道有多少个数的数字之和恰好为n。例如,当n=2时,有5个数:11、14、41、44、2。
故 F(n) = F(n - 1) + F(n - 2) + F(n - 3) + F(n - 4) = 2 * F(n - 1) + F(n - 2) + F(n - 3)。
边界条件: F(1) = 2, F(2) = 5, F(3) = 13。
#include<string>#include<iostream>#include<vector>#include<algorithm>using namespace std;vector<string> v;string add(string a, string b){ string s; reverse(a.begin(), a.end()); reverse(b.begin(), b.end()); int i = 0; int m, k = 0; while(a[i] && b[i]) { m = a[i] - '0' + b[i] - '0' + k; k = m / 10; s += (m % 10 + '0'); i++; } if(i == a.size()) { while(i != b.size()) { m = k + b[i] - '0'; k = m / 10; s += m % 10 + '0'; i++; } if(k) s += k + '0'; } else if(i == b.size()) { while(i != a.size()) { m = k + a[i] - '0'; k = m / 10; s += m % 10 + '0'; i++; } if(k) s += k + '0'; } reverse(s.begin(), s.end()); return s;}void solve(){ v.push_back("0"); v.push_back("2"); v.push_back("5"); v.push_back("13"); string s; for(int i = 4; ; i++) { s = add(v[i-1], v[i-1]); s = add(v[i-2], s); s = add(v[i-3], s); v.push_back(s); if(v[i].size() > 1001) break; }}int main(){ solve(); int n; int Size = v.size(); while(cin >> n) { cout << v[n] << endl; } return 0;}
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