[Codeforces Round #155 (Div. 2)]A. Cards with Numbers
2013-01-18 21:22
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地址:http://codeforces.com/contest/254/problem/A
这题昨天就在想,想复杂了一些
一个5000的数组用来存数字的位数,一个400000x2的二维数组用来存答案(400000稍大了些)
读每个数第一次读入时,先将其位置存到5000数组里,第二次读入时,一起转到二维数组中,然后5000数组相应位置清零
所有数据读完后,检查5000数组,全为零则可以输出答案,否则输出-1
这题昨天就在想,想复杂了一些
一个5000的数组用来存数字的位数,一个400000x2的二维数组用来存答案(400000稍大了些)
读每个数第一次读入时,先将其位置存到5000数组里,第二次读入时,一起转到二维数组中,然后5000数组相应位置清零
所有数据读完后,检查5000数组,全为零则可以输出答案,否则输出-1
#include<stdio.h> int n,a[5001],ans[400000][2]; int main() { int i,in,j=0,flag=0; freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); scanf("%d",&n); for(i=1;i<=2*n;i++) { scanf("%d",&in); if(0==a[in]) a[in]=i; else { ans[j][0]=a[in]; ans[j][1]=i; j++; a[in]=0; } } for(i=1;i<=5000;i++) { if(a[i]!=0) {flag=1;break;} } if(flag) printf("-1\n"); else { while(j>0) { printf("%d %d\n",ans[j-1][0],ans[j-1][1]); j--; } } return 0; }
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