Ural 1915 - Reconstruction of Bygones 模拟
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题意是说当输入为正数时~压入栈..当输入为0时将所有元素复制一遍进去..当输入为-1时.输出栈顶并弹出...
练习赛的时候看范围就不敢做了...其实..注意当栈内的元素比较多了..就不需要复制了...卡住条件..直接模拟就行...
Program:
#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<algorithm>#include<queue>#include<stack>#define ll long long #define oo 1000000007#define MAXN 1000005using namespace std; int n,mystack[MAXN<<1];int main(){ int t,i,top; while (~scanf("%d",&n)) { top=0; for (t=n;t>=1;t--) { int x; scanf("%d",&x); if (x>0) mystack[++top]=x; if (x==-1) printf("%d\n",mystack[top--]); if (!x) { if (top>t) continue; //没必要复制了 for (i=1;i<=top;i++) mystack[top+i]=mystack[i]; top*=2; } } } return 0;}
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