Jonathan is fighting against DIO's Vampire minions. There are n of them with strengths a1,a2,…,an. Denote (l,r) as the group consisting of the vampires with indices from l to r. Jonathan realizes that the strength of any such group is in its weakest link, that is, the bitwise AND. More formally, the strength level of the group (l,r) is defined as f(l,r)=al&al+1&al+2&…&ar. Here, & denotes the bitwise AND operation. Because Jonathan would like to defeat the vampire minions fast, he will divide the vampires into contiguous groups, such that each vampire is in exactly one group, and the sum of strengths of the groups is minimized. Among all ways to divide the vampires, he would like to find the way with the maximum number of groups. Given the strengths of each of the n vampires, find the maximum number of groups among all possible ways to divide the vampires with the smallest sum of strengths. Input The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer n (1≤n≤2⋅105) — the number of vampires. The second line of each test case contains n integers a1,a2,…,an (0≤ai≤109) — the individual strength of each vampire. The sum of n over all test cases does not exceed 2⋅105. Output For each test case, output a single integer — the maximum number of groups among all possible ways to divide the vampires with the smallest sum of strengths.c++实现
时间: 2024-04-07 14:30:28 浏览: 21
您的问题是如何使用C++实现解决这个问题。
以下是使用C++编写的解决方案:
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> strengths(n);
for (int i = 0; i < n; i++) {
cin >> strengths[i];
}
// Sort the strengths in non-decreasing order
sort(strengths.begin(), strengths.end());
int minStrength = strengths[0];
int groupCount = 1;
// Count the number of groups with the smallest strength
for (int i = 1; i < n; i++) {
if (strengths[i] == minStrength) {
groupCount++;
} else {
break;
}
}
cout << groupCount << endl;
}
return 0;
}
```
这段代码首先读取测试用例的数量t。然后,对于每个测试用例,它读取吸血鬼的数量n和每个吸血鬼的强度。然后,它对吸血鬼的强度进行排序,并找到最小的强度minStrength。接下来,它遍历强度数组并计算具有最小强度的组数groupCount。最后,将groupCount输出到标准输出。
请注意,这只是一种可能的解决方案,并且可能不是最优解。还有其他方法可以解决这个问题。
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