agc007B - Construct Sequences(构造)

题意

题目链接

给出一个$1-N$的排列$P$，构造两个数组$a, b$满足

Sol

发现我的水平也就是能做一做0-699的题。。。。

直接构造两个等差数列$a, b$，公差为$20000$

然后从小到大枚举$p$，让考前的$a$减去一个较大的数就行了。。

#include<bits/stdc++.h>
#define LL long long  
using namespace std;
const int MAXN = 20001, base = 20009;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, a[MAXN], b[MAXN];
int main() {
    N = read();
    for(int i = 1; i <= N; i++) a[i] = base * i;
    for(int i = 1; i <= N; i++) b[i] = base * (N - i + 1);
    for(int i = 1; i <= N; i++) {
        int x = read();
        a[x] -= (N - i + 1);
    }
    for(int i = 1; i <= N; i++) printf("%d ", a[i]); puts("");
    for(int i = 1; i <= N; i++) printf("%d ", b[i]);
    return 0;
}