Project Euler:problem 104
2017-07-03 20:04 mo_0820 阅读(169) 评论(0) 收藏 举报题目链接:http://pe-cn.github.io/104/
# include <stdio.h> # include <stdlib.h> void sort(int x,int y,char *a) { int xx=x,yy=y; char k=a[x]; if(x>=y) return ; while(xx!=yy) { while(xx<yy&& a[yy]>=k)yy--; a[xx]=a[yy]; while(xx<yy&& a[xx]<=k)xx++; a[yy]=a[xx]; } a[xx]=k; sort(x,xx-1,a); sort(xx+1,y,a); } int main() { char *a, *b, *c ,d[9]; int i,j,k=0,n,N; while((a=(char *)malloc(sizeof(char)*100))==NULL); while((b=(char *)malloc(sizeof(char)*100))==NULL); while((c=(char *)malloc(sizeof(char)*100))==NULL); for(i=0;i<100;i++) { a[i]=0; b[i]=0; c[i]=0; } N=100; a[0]=1; b[0]=1; n=0; for(i=3;;i++) { for(j=0;j<=n;j++) c[j]=a[j]+b[j]; for(j=0;j<=n;j++) if(c[j]>=10) { c[j+1]+=c[j]/10; c[j]%=10; } if(c[n+1]>0) n+=1; if(n>10) { for(j=0;j<9;j++) d[j]=c[j]; sort(0, 8, d); for(j=0;j<9;j++) if(d[j]!=j+1) break; if(j==9) { printf("last nine %d\n",i); k=1; } for(j=n;j>n-9;j--) d[n-j]=c[j]; sort(0, 8, d); for(j=0;j<9;j++) if(d[j]!=j+1) break; if(j==9) { printf("first nine %d\n",i); k+=1; } if(k==2) { printf("last and first nine %d\n",i); break; } else k=0; } if(n==N-1) { N+=100; free(a); a=NULL; while((a=(char *)malloc(sizeof(char)*N))==NULL); for(j=0;j<N-100;j++) a[j]=b[j]; for(;j<N;j++) a[j]=0; free(b); b=NULL; while((b=(char *)malloc(sizeof(char)*N))==NULL); for(j=0;j<N-100;j++) b[j]=c[j]; for(;j<N;j++) b[j]=0; free(c); c=NULL; while((c=(char *)malloc(sizeof(char)*N))==NULL); for(j=0;j<N;j++) c[j]=0; } else { for(j=0;j<=n;j++) { a[j]=b[j]; b[j]=c[j]; c[j]=0; } } } return 0; }
部分结果:
first nine 308563
first nine 309329
last nine 309354
last nine 309998
first nine 310064
last nine 310229
first nine 311434
first nine 312461
last nine 312971
last nine 314328
first nine 315817
first nine 315965
first nine 317784
first nine 320465
last nine 320556
first nine 322658
first nine 325047
last nine 327712
last nine 328733
last nine 329468
first nine 329468
last and first nine 329468
Press any key to continue
浙公网安备 33010602011771号