42 lines
1.0 KiB
Markdown
42 lines
1.0 KiB
Markdown
|
||
// 计算有多少种方式
|
||
// 完全背包 for循环的顺序啊，先是哪个for 后是哪个for定不下来
|
||
|
||
排列
|
||
```
|
||
class Solution {
|
||
public:
|
||
int change(int amount, vector<int>& coins) {
|
||
int dp[50001] = {0};
|
||
dp[0] = 1;
|
||
for (int i = 0; i <= amount; i++) {
|
||
for (int j = 0; j < coins.size(); j++) { // 这是组合把？？？
|
||
if (i - coins[j] >= 0) dp[i] += dp[i - coins[j]];
|
||
}
|
||
for (int j = 0; j <= amount; j++) {
|
||
cout << dp[j] << " ";
|
||
}
|
||
cout << endl;
|
||
}
|
||
return dp[amount];
|
||
}
|
||
};
|
||
```
|
||
|
||
这个才是组合，本题的题解，
|
||
```
|
||
class Solution {
|
||
public:
|
||
int change(int amount, vector<int>& coins) {
|
||
int dp[50001] = {0};
|
||
dp[0] = 1;
|
||
for (int i = 0; i < coins.size(); i++) { // 一个钱币只在序列里出现一次
|
||
for (int j = 0; j <= amount; j++) {
|
||
if (j - coins[i] >= 0) dp[j] += dp[j - coins[i]];
|
||
}
|
||
}
|
||
return dp[amount];
|
||
}
|
||
};
|
||
```
|