和之前个回溯法的各个总和串起来一波，本题用回溯稳稳的超时

```
class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target + 1, 0);
        dp[0] = 1;
        for (int i = 0; i <= target; i++) {
            for (int j = 0; j < nums.size(); j++) {
                if (i - nums[j] >= 0) {
                    dp[i] += dp[i - nums[j]];
                }
            }
        }
        return dp[target];
    }
};
```


C++测试用例有超过两个树相加超过int的数据
超限的情况 


一些题解会直接用ull  usigned long long

java 也是四个字节，理论上没有差别，可能后台java和C++测试用例不同，bug.....
```
class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target + 1, 0);
        dp[0] = 1;
        for (int i = 0; i <= target; i++) {
            for (int num : nums) {
                if (i - num >= 0 && dp[i] < INT_MAX - dp[i - num]) {
                    dp[i] += dp[i - num];
                }
            }
        }
        return dp[target];
    }
};
```