2019NC#1

LINK

B Integration

题意：

给定$a_1,a_2,...,a_n$, 计算 $$frac{1}{π}int_{0}^{infty}frac{1}{prodlimits_{i=1}^{n}(a_i^2+x^2)}dx$$ 在mod（1E9+7）意义下的答案。

思路：

裂项化乘为和的方法

可以得到

$$frac{1}{2}sum_{i=1}^n quad  frac{1}{prod_{j=1,j e i}^n quad a_j^2 - a_i^2} quad frac{1}{a_i}$$

参考：https://www.cnblogs.com/Dillonh/p/11209476.html

 

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;


template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(ll &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

/**********showtime************/
            const int maxn = 1e3+9;
            int a[maxn];
            ll ksm(ll a, ll b) {
                ll res = 1;
                while(b > 0) {
                    if(b & 1) res = res * a % mod;
                    a = a * a % mod;
                    b = b >> 1;
                }
                return res;
            }
int main(){
            int n;  
            while(~scanf("%d", &n)){
                for(int i=1; i<=n; i++) scanf("%d", &a[i]);
                ll sum = 0;
                for(int i=1; i<=n; i++) {

                    ll tmp = 1;
                    for(int j=1; j<=n; j++) {
                        if(i == j) continue;
                        ll s = (1ll*a[j] * a[j]%mod - 1ll*a[i]*a[i]%mod + mod)%mod;
                        tmp = tmp * ksm(s, mod-2) % mod;
                    }
                    tmp = tmp * ksm(a[i], mod-2) % mod;
                    sum = (sum + tmp )% mod;
                }
                sum = sum * ksm(2, mod-2) % mod;
                printf("%lld
", sum);
            }
            return 0;
}

C Euclidean Distance

贪心或者拉格朗日乘子法

D Parity of Tuples

fwt

G Substrings 2

字符串

H XOR

线性基

I Points Division

DP，线段树，折线