CF #329 C

C题我还以为是拉格朗日插值。。。

其实可以想象到，必须有这样一个函数，经过某一点时，其它圆相关的函数要为0。

于是，可以构造这样的一个函数，对于x有 （x/2）*(1-abs(t-i)+abs(1-abs(t-i)))

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>

using namespace std;

#define LL __int64
#define PB push_back
#define P pair<int, int>
#define X first
#define Y second
const int N = 55;
P a[N];

int main() {
    int n;
    scanf("%d", &n);
    int r;
    for(int i = 0; i < n; ++i) {
        scanf("%d%d%d", &a[i].X, &a[i].Y, &r);
    }
    string left = "(((1-abs((t-", mid = ")))+abs((abs((t-", right = "))-1)))";
    string ans1 = "";
    string ans2 = "";
    for(int i = 0; i < n - 1; ++i) {
        ans1 += "(";
        ans2 += "(";
    }
    for(int i = 0; i < n; ++i) {
        string num = to_string(i);
        ans1 += left + num + mid + num + right + "*" + to_string(a[i].X / 2) + ")";
        ans2 += left + num + mid + num + right + "*" + to_string(a[i].Y / 2) + ")";
        if(i != 0) ans1 += ")", ans2 += ")";
        if(i != n - 1) ans1 += "+", ans2 += "+";
    }
    cout<<ans1<<endl;
    cout<<ans2<<endl;
    return 0;
}