ZOJ1246 POJ1472 HDU1468 Instant Complexity

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很烦躁的题, 不管是输入还是输出, 都给人奔溃的感觉

用递归吧, 设个临时数组保存子循环的时间复杂度

#ifdef _MSC_VER#define DEBUG#define _CRT_SECURE_NO_DEPRECATE#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define MAX  20using namespace std;int order[1000][2];void func(const int &beg,const int &end,int *num){for(int i=beg;i<end;){if(order[i][0]==3){num[0]+=order[i][1];++i;}else if(order[i][0]==2){int temp[20];memset(temp,0,sizeof(temp));int sum=1;int t=i+1;while(sum){if(order[t][0]==2)++sum;else if(order[t][0]==1)--sum;++t;}func(i+1,t-1,temp);if(order[i][1]==INT_MAX)for(int j=0;j<MAX-1;++j)num[j+1]+=temp[j];elsefor(int j=0;j<MAX;++j)num[j]+=temp[j]*order[i][1];i=t;}}}int main(void){#ifdef DEBUG    freopen("../stdin.txt","r",stdin);  freopen("../stdout.txt","w",stdout); #endif  string str,cost;int ncases;scanf("%d",&ncases);  cin>>str;for(int nc=1;nc<=ncases;++nc){int n=0;    int sum=1;while(sum){if(str=="BEGIN" )order[n++][0]=0;else if(str=="END")      {        order[n++][0]=1;         --sum;      }else {cin>>cost;int temp;if(cost=="n")temp=INT_MAX;elsetemp=atoi(cost.c_str());if(str=="LOOP")        {order[n][0]=2;          ++sum;        }else        {order[n][0]=3;        }order[n++][1]=temp;}      cin>>str;}int ans[MAX];memset(ans,0,sizeof(ans));func(1,n-1,ans);printf("Program #%d\nRuntime =",nc);    bool flag=false;    for(int i=MAX-1;i>1;--i)      if(ans[i]>1)      {        printf("%c%d*n^%d",flag?'+':' ',ans[i],i);        flag=true;      }      else if(ans[i]==1)      {        printf("%cn^%d",flag?'+':' ',i);        flag=true;      }      if(ans[1]>1)      {        printf("%c%d*n",flag?'+':' ',ans[1]);        flag=true;      }      else if(ans[1])      {        printf("%cn",flag?'+':' ');        flag=true;      }      if(ans[0])      {        printf("%c%d",flag?'+':' ',ans[0]);        flag=true;      }      if(!flag)        printf(" 0");      printf("\n\n");}  return 0;}