Codeforces Round #332 (Div. 2) B. Spongebob and Joke
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题意:若第三行的数在第二行的数中出现过,并且全部为出现1次的,则输出possible,并输出其对应的在第二行中的位置。若出现过两次,则输出Ambiguity,否则,输出Impossible.
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxm=1e5+10;int a[maxm];int f[maxm];int b[maxm];int vis[maxm];int vit[maxm];int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF) { memset(vis,0,sizeof(vis)); memset(vit,0,sizeof(vit)); for(int i=1; i<=n; i++) { scanf("%d",&f[i]); vis[f[i]]=i; vit[f[i]]++; } for(int i=1; i<=m; i++) { scanf("%d",&b[i]); } int cnt=0; int ok1=0,ok2=0; for(int i=1; i<=m; i++) { if(!vit[b[i]]) { ok1=1; } else if(vit[b[i]]>1) { ok2=1; } else { a[cnt++]=vis[b[i]]; } } if(ok1) { printf("Impossible\n"); } else if(ok2) { printf("Ambiguity\n"); } else { printf("Possible\n"); for(int i=0; i<m; i++) { if(i==0) printf("%d",a[i]); else printf(" %d",a[i]); } printf("\n"); } } return 0;}
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