3-idiots

思路

多项式卷积的经典应用了：

看样例一:

4
1 3 3 4

我们用多项式系数表示得到一个与$a_i$有关的多项式：$01021$，

也就说明原序列有0个1，1个1，0个2，2个3，1个4，

这个多项式卷积之后得到：0 0 1 0 4 2 4 4 1

也就是可重复使用上面的两个数可以得到1个2$\dots$

由此我们就确定了三角形的两条边了，但是因为可重复使用+有序，所以我们要对上面的数值做一点点改变。

for(int i = 1; i <= n; i++) {num[x[i] + x[i]]--;
}//去除重复使用
for(int i= 0; i < lim; i++) {num[i] >>= 1;
}//把序列变成有序状态

接下来就是找出合法的组合方案了。

我们枚举最大的边，把不合法的方案都给减去即可。

代码

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const double pi = acos(-1.0);const int N = 1e6 + 10;struct Complex {double r, i;Complex(double _r = 0, double _i = 0) : r(_r), i(_i) {}}a[N];Complex operator + (const Complex & a, const Complex & b) {return Complex(a.r + b.r, a.i + b.i);
}Complex operator - (const Complex & a, const Complex & b) {return Complex(a.r - b.r, a.i - b.i);
}Complex operator * (const Complex & a, const Complex & b) {return Complex(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r);
}int r[N], x[N], n, m;ll num[N];void fft(Complex * f, int lim, int rev) {for(int i = 0; i < lim; i++) {if(r[i] < i) {swap(f[i], f[r[i]]);}}for(int i = 1; i < lim; i <<= 1) {Complex wn = Complex(cos(pi / i), rev * sin(pi / i));for(int p = i << 1, j = 0; j < lim; j += p) {Complex w = Complex(1, 0);for(int k = 0; k < i; k++, w = w * wn) {Complex x = f[j + k], y = w * f[i + j + k];f[j + k] = x + y, f[i + j + k] = x - y;}}}if(rev == -1) {for(int i = 0; i < lim; i++) {f[i].r /= lim;}}
}void get_r(int lim) {for(int i = 0; i < lim; ++i) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T, lim;scanf("%d", &T);while(T--) {scanf("%d", &n);memset(num, 0, sizeof num);for(int i = 1; i <= n; i++) {scanf("%d", &x[i]);num[x[i]]++;}sort(x + 1, x + 1 + n);lim = 1;while(lim <= (2 * x[n])) {lim <<= 1;}for(int i = 0; i <= x[n]; i++) {a[i] = Complex(num[i], 0);}for(int i = x[n] + 1; i < lim; i++) {a[i] = Complex(0, 0);}get_r(lim);fft(a, lim, 1);for(int i = 0; i < lim; i++) {a[i] = a[i] * a[i];}fft(a, lim, -1);for(int i = 0; i < lim; i++) {num[i] = int(a[i].r + 0.5);}for(int i = 1; i <= n; i++) {num[x[i] + x[i]]--;}for(int i= 0; i < lim; i++) {num[i] >>= 1;}ll tot = 1ll * n * (n - 1) * (n - 2) / 6, ans = tot;for(int i = 1, k = 1; i < lim; i++) {if(num[i]) {while(k <= n && i > x[k]) k++;if(k == n + 1) break;ans -= num[i] * (n - k + 1);}}printf("%.7f\n", 1.0 * ans / tot);}return 0;
}

C - Golf Bot（附赠裸题）

思路

多项式相乘一下，然后判断哪些值出现了，再记个数就行，FFT裸题。

代码

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const double pi = acos(-1.0);const int N = 1e6 + 10;struct Complex {double r, i;Complex(double _r = 0, double _i = 0) : r(_r), i(_i) {}}a[N];Complex operator + (const Complex & a, const Complex & b) {return Complex(a.r + b.r, a.i + b.i);
}Complex operator - (const Complex & a, const Complex & b) {return Complex(a.r - b.r, a.i - b.i);
}Complex operator * (const Complex & a, const Complex & b) {return Complex(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r);
}int r[N], k[N], num[N], d[N], n, m, ans, lim;void fft(Complex * f, int lim, int rev) {for(int i = 0; i < lim; i++) {if(r[i] < i) {swap(f[i], f[r[i]]);}}for(int i = 1; i < lim; i <<= 1) {Complex wn = Complex(cos(pi / i), rev * sin(pi / i));for(int p = i << 1, j = 0; j < lim; j += p) {Complex w = Complex(1, 0);for(int k = 0; k < i; k++, w = w * wn) {Complex x = f[j + k], y = w * f[i + j + k];f[j + k] = x + y, f[i + j + k] = x - y;}}}if(rev == -1) {for(int i = 0; i < lim; i++) {a[i].r /= lim;}}
}void get_r(int lim) {for(int i = 0; i < lim; ++i) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d", &n);for(int i = 1; i <= n; i++) {scanf("%d", &k[i]);num[k[i]]++;}sort(k + 1, k + 1 + n);lim = 1;while(lim < k[n] * 2) lim <<= 1;for(int i = 0; i < lim; i++) {a[i] = Complex(num[i], 0);}get_r(lim);fft(a, lim, 1);for(int i = 0; i < lim; i++) {a[i] = a[i] * a[i];}fft(a, lim, -1);for(int i = 0; i < lim; i++) {num[i] += int(a[i].r + 0.5);}scanf("%d", &m);for(int i = 1; i <= m; i++) {int x;scanf("%d", &x);if(num[x]) {num[x] = 0;ans++;}}printf("%d\n", ans);return 0;
}