Codeforces Round #310 (Div. 2)--B

http://codeforces.com/problemset/problem/556/B

题意：给定n个数字且都小于n，然后每次循环第2k+1个数字+1，第2k个数字减一，k=0,1,2...n/2.  问最后能不能使n个数字刚好为排列为0,1,2，....n-1.

题解：简单模拟就好了，循环次数设为n-1，因为n次循环后回到原来的排列。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
using namespace std;
typedef long long ll;
const int maxn=1010;
int n,a[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    int t=1,sum=0;
    while(t<=n){
        for(int i=0;i<n;i++){
            if(a[i]==i) sum++;
            if(i%2==0){
                if(a[i]==n-1) a[i]=0;
                else a[i]+=1;
            }
            else{
                if(a[i]==0) a[i]=n-1;
                else a[i]-=1;
            }
        }
        if(sum==n) break;
        else sum=0;
        t++;
    }
    if(sum==n) printf("Yes
");
    else printf("No
");
    return 0;
}