lightoj 1374 Confusion in the Problemset

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题目就是说有n页纸，有n个页码，由于存在某些问题导致页码和平时的不一样，这里每页的页码指的是这页前面或者后面还有多少页，问给定的这些页码是否可以排列出来满足上面条件的序列。主意的是页码的范围是<=10^6,n <= 10000。

ps：开始想错了，一位需要将大于n或者小于0的数进行hash，结果wa了三发没懂为啥，，，太蠢了，题目没懂

点击打开题目链接

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 10010;int A[maxn];int t,icase = 0;int main(){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    int n;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        memset(A, 0,sizeof A);        int x;        for (int i = 1;i <= n;++i){            scanf("%d",&x);            if (x > n || x < 0) continue;            A[x] += 1;        }        bool flag = true;        for (int i = 1;i <= n;++i){            int first = i - 1;//指示前面有first页            int second = n - i;//指示后面有second页            if (A[first] > 0){                A[first]--;            }else if (A[second] > 0){                A[second]--;            }            else{                flag = false;                break;            }        }        printf("Case %d: ", ++icase);        if (flag) printf("yes\n");        else printf("no\n");    }    return 0;}

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