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day2

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好想长脑子 , 2025-03-27 19:53:46 · 广东 , 所有人可见 , 阅读 3

最小战斗力差距.jpg
`#include[HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 100010;
typedef long long LL;
using namespace std;

LL w[N];

int main(){

int n;
cin >> n;

for(int i = 0; i < n; i ++) cin >> w[i];
//排序将a、b分成两组 从数组中所取 可理解为左侧a的最大 右侧为b的最小
sort(w, w + n);
LL res = 1e9 + 10;
for(int i = 1; i <= n; i ++)
{
    res = min(res, abs(w[i] - w[i - 1]));
}



printf("%ld", res);
return 0;

} `

谈判.jpg
`//小根堆#include[HTML_REMOVED]
//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 100010;
typedef long long LL;
using namespace std;

LL w[N];

int main(){

int n;
cin >> n;
//priority_queue<数据类型， 容器类型， greater构造成小根堆> pq；

priority_queue<int, vector<int>, greater<int>> pq;
int t = -1;
for(int i = 0; i < n; i ++) 
{
    cin >> t;
    pq.push(t); 
}
LL res = 0;
while(pq.size() > 1)
{
    int sum1 = pq.top();
    pq.pop();
    int sum2 = pq.top();
    pq.pop();
    int sum3 = sum1 + sum2;
    pq.push(sum3);
    res += sum3;
}


cout << res;
//printf("%ld", res);
return 0;

} `

纪录片分组.jpg

////暴力实现
`#include[HTML_REMOVED]
//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 30010;
typedef long long LL;
using namespace std;

int main(){
int w;
int p[N];
cin >> w;
int n;
cin >> n;

for(int i = 0; i < n; i ++) cin >> p[i];
//排序 不排序不满足从大开始实现
sort(p, p + n);
for(int i = n - 1; i >= 0; i --)
{   
    int flag = 0;
    if(p[i] == -1) continue;

    for(int j = i - 1; j >= 0; j --)
    {   
        if(p[j] == -1) continue;

        if(p[i] + p[j] <= w)
        {

            p[i] = -1; 
            p[j] = -1;
            flag = 1;
            break;
        }
    }
}
int sum = 0;
int sum1 = 0;
for(int i = 0; i < n; i ++)
{
    if(p[i] < 0) sum1++;
    else sum++;
}
cout << sum + sum1 / 2;
return 0;

}
////双指针

include[HTML_REMOVED]

//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 30010;
typedef long long LL;
using namespace std;

int main(){
int w;
int p[N];
cin >> w;
int n;
cin >> n;

for(int i = 0; i < n; i ++) cin >> p[i];

sort(p, p + n);
int sum = 0;
int l = 0, r = n - 1;
while(l <= r)
{
    if(p[l] + p[r] <= w)
    {
        l ++; r --;
        sum ++;
    }
    else
    {
        r --;
        sum ++;
    }
}
cout << sum;
return 0;

}
`
回文判定.jpg

//双指针
`#include[HTML_REMOVED]
//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 30010;
typedef long long LL;
using namespace std;

int main(){
string s;
cin >> s;

int l = 0, r = s.length() - 1;
int flag = 1;

while(l <= r)
{
    if(s[l] != s[r]) 
    {
        flag = 0;
        break;
    }
    else
    {
        l ++; r --;
    }
    //if(l == r) break;
}

if(flag) cout << "Y";
else cout << "N";
return 0;

} `

美丽的区间.jpg

`#include[HTML_REMOVED]
//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 1e5 + 10;
typedef long long LL;
using namespace std;

int main(){
int n, s;
cin >> n >> s;

int a[N], b[N];
int len = 1e5 + 10;
int ans = len;
for(int i = 1; i <= n; i ++) cin >> a[i];
b[0] = 0;
for(int i = 1; i <= n; i ++) b[i] = b[i - 1] + a[i];
int j = 1;
for(int i = 1; i <= n; i ++)
{
    if(j < i) j = i;
    while(b[j] - b[i - 1] < s && j <= n)
    {   
        j ++;
    }

    if(j <= n)  ans = min(ans, j - i + 1);

}

if(ans == len) cout << 0;
else cout << ans << endl;


return 0;

} `

`//二进制求1的个数
#include[HTML_REMOVED]
//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 1e5 + 10;
typedef long long LL;
using namespace std;

//计算最右边位于1的位置模板适用
int lowbits(int x)
{
return x & (-x);
}

int main(){
int n;
cin >> n;
int b = 1;

LL cnt = 0;


while(n)
{
    n -= lowbits(n);
    cnt ++;
}
cout << cnt;


return 0;

} `

异或森林.png

`#include[HTML_REMOVED]
//#include [HTML_REMOVED]

include[HTML_REMOVED]

include [HTML_REMOVED]

const int N = 1e4 + 10;
typedef long long LL;
using namespace std;

int lowbits(int x)
{
return x & (-x);
}
///////如何正确的判断该数是否为完全平方数
//根据数论知识：一个正整数因数个数为奇数当且仅当它是完全平方数（0 的情况需特殊处理），因此因数个数为偶数通常等价于该数不是一个完全平方数（题目可能对 0 也有规定）。
int is(int a)
{
int r = (int)(sqrt(a) + 1e-9);
return (r * r == a);
}
int a[N];

int main(){
int n;
cin >> n;

for(int i = 1; i <= n; i ++) cin >> a[i];
int cnt = 0;
for(int i = 1; i <= n; i ++)
{   
    int t = 0;

    for(int j = i; j <= n; j ++)
    {
        //cout << t << " ";
        t ^= a[j];
        if(!is(t) && t != 0) cnt ++;

        //cout << t << " ";
    }

}
cout << cnt;

return 0;

} `

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