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;
- }
经验首页