div2#730题解

div#730把我打自闭了
害！看开了赛后好好补题才是关键，没有翻不过的高山，只要努力成功会向我走来的！
大家都一样嗷~

A

怎么说呢这道题是压倒我好心情的重要稻草
淦·好题
这道题思维首先第一个坑你得开long long
第二个不能模拟
wa了3回。。。人间惨案不说了不说了

#include <algorithm>
#include <iostream>
#include <cstring>using namespace std;
typedef long long LL;
LL a, b,k;
LL gcd(LL a, LL b)
{return b?gcd(b, a % b):a;
}int main()
{cin >> k;while (k --){cin >> a >> b;if (a < b) swap(a, b);LL t = a - b;if (t == 0) cout << 0 << " " << 0 << endl;else{cout << a - b << " ";LL a1 = a % (a - b);LL b1 = b % (a - b);if (t - a1 > a1) cout << a1 << endl;else cout << t - a1 << endl;}}return 0;
}

B

B题挺水的但是阿巴阿巴

#include <algorithm>
#include <iostream>
#include <cstring>using namespace std;const int N = 2e5 + 10;
typedef long long ll;
ll a[N];
int n;int main()
{int t;cin >> t;while (t -- ){cin >> n;ll sum = 0;for (int i = 1; i <= n; i ++ ){cin >> a[i];sum += a[i];}if (n == 1) cout << 0 << endl;else if (sum % n == 0) {cout << 0 << endl;}else {ll k = sum % n;ll r = n - k;cout << (sum % n) * r << endl;}}return 0;
}

C

c题比赛的时候好多人没有读懂题目什么意思
就是和高中学的概率一样的其实
可以用好多方法去完成这里提供一种dfs

/*
一定要注意
"the item is no longer a valid item for all the further draws"
如果一个概率等于0之后就不在加了一值保持0的状态
*/
#include <algorithm>
#include <iostream>
#include <cstring>using namespace std;double c,m,p,v,t,sum;
const int N = 1e-6;void dfs(double c, double m, double p, double v, double a, int n)
{sum += n * a * p;if (c <= N && m <= N) return;if (c > N) {double x = min(c, v);if (m > N) {//cout << a * c << endl;dfs(c - x, m + x / 2, p + x / 2, v, a * c, n + 1);}else {//cout << a * c << endl;dfs(c - x, 0, p + x, v, a * c, n + 1);}}if (m > N){double x = min(m, v);if (c > N) {//cout << a * m << endl;dfs(c + x / 2, m - x, p + x / 2, v, a * m, n + 1);}else {//cout << a * m << endl;dfs(0, m - x, p + x, v, a * m, n + 1);}}
}int main()
{int t;scanf("%d",&t);while (t -- ){scanf("%lf%lf%lf%lf",&c,&m,&p,&v);sum = 0;dfs(c, m, p, v, 1.0, 1);printf("%.12f\n",sum);}return 0;
}

D

交互题位运算

#include <algorithm>
#include <iostream>using namespace std;int r, n, k, y;int main()
{int t;cin >> t;while (t -- ){cin >> n >> k;int s = 0;for (int i = 0; i < n; i ++){y = i ^ s;cout << y << endl;fflush(stdout);s = i;cin >> r;if (r == 1) break;}}return 0;
}

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div2#730题解

A

B

C

D

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