思路
动态规划思想,把A和B歌单编程变成数组:[A,A,A,B,B,B],然后,如果需要长度为k的歌组合f(n,k), n表示前n个数,k表示歌单长度为k. 则,f(n,k)=f(n-1,k)+f(n,k-len[n-1]);k-len[n-1]表示数组中第n-1个数使用的情况下,还需要长度为k-len[n-1]个歌单的数量。
code
#include "stdafx.h"
#include <iostream>
#include <vector>
using namespace std;
int main()
{
const int MOD = 1000000007;
int n;
cin >> n;
int l1, l2, n1, n2;
cin >> l1 >> n1 >> l2 >> n2;
vector<int> len(n1 + n2);
for (int i = 0; i < n1; ++i) {
len[i] = l1;
}
for (int i = n1; i < n1 + n2; ++i) {
len[i] = l2;
}
vector<vector<int>> results(n1 + n2, vector<int>(n + 1));
for (int i = 0; i < n1 + n2; ++i) {
results[i][0] = 1;
}
results[0][len[0]] = 1;
for (int i = 1; i < n1 + n2; ++i) {
for (int j = 1; j <= n; ++j) {
if (j - len[i] >= 0) {
results[i][j] += results[i - 1][j - len[i]];
}
results[i][j] += results[i - 1][j];
results[i][j] %= MOD;
}
}
cout << results[n1 + n2 - 1][n];
return 0;
}