地址:http://codeforces.com/contest/1054/problem/C
这道题是简单题就是我没有逆向思维,思维不会变通,当时想到,如果给的L,R数组是正确的话,那么答案就是n - L[i] - R[i],然后就一直在想如何判断给的数组一定输出YES,其实可以先减得出结果数组,然后再判断是否可以得出L,R数组,如果不能,输出NO,否则输出YES和结果数组。。。。。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
const int N = 1005;
int L[N];
int R[N];
int res[N];
int main()
{
int n;
scanf("%d",&n);
for(int i = 1;i <= n;++i) scanf("%d",&L[i]);
for(int i = 1;i <= n;++i) scanf("%d",&R[i]);
for(int i = 1;i <= n;++i) res[i] = n - L[i] - R[i];
for(int i = 1;i <= n;++i){
for(int j = i + 1;j <= n;++j){
if(res[j] > res[i]){
R[i]--;
}
if(res[j] < res[i]){
L[j]--;
}
}
}
for(int i = 1;i <= n;++i){
if(L[i] != 0 || R[i] != 0){
printf("NO\n");
return 0;
}
}
printf("YES\n");
for(int i = 1;i <= n;++i){
printf("%d ",res[i]);
}
printf("\n");
return 0;
}