1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 | #include <stdio.h> #include <string.h> int compare( int *, int *); void swap( int * , int * ); void deal( int *, int *, int , int ); int main() { char s1[100],s2[100]; int oper1[100]={0},oper2[100]={0},m,n,c,k,i; gets (s1); gets (s2); m= strlen (s1); n= strlen (s2); (m<n? k=n : k=m); c=k; for (i=0;i<m;i++) oper1[k--]=s1[m-1-i]- '0' ; for (k=c,i=0;i<n;i++) oper2[k--]=s2[n-1-i]- '0' ; if (compare(oper1,oper2)==3) puts ( "0" ); if (compare(oper1,oper2)==1)deal(oper1,oper2,c,k); if (compare(oper1,oper2)==2) { swap(oper1,oper2); printf ( "-" ); deal(oper1,oper2,c,k); } } void deal( int * oper1, int * oper2, int c, int k) { for ( int i=c;i>=0;i--) { oper1[i]-=oper2[i]; if (oper1[i]<0) { oper1[i]+=10; oper1[i-1]--; } } for ( int i=0;i<c;i++) if (oper1[i]==0&&oper1[i+1])k=i+1; for ( int i=k;i<=c;i++) printf ( "%d" ,oper1[i]); } void swap( int * oper1, int * oper2) { for ( int i=0;i<100;i++) {oper1[i]=oper1[i]+oper2[i]; oper2[i]=oper1[i]-oper2[i]; oper1[i]=oper1[i]-oper2[i]; } } int compare( int * oper1, int * oper2) { for ( int i=0;i<100;i++) { if (oper1[i]>oper2[i]) return 1; //操作数oper1大 if (oper1[i]<oper2[i]) return 2; //操作数oper2大 } return 3; //操作数oper1和操作数oper2一样大 } |
0 分
0 人评分
C语言网提供由在职研发工程师或ACM蓝桥杯竞赛优秀选手录制的视频教程,并配有习题和答疑,点击了解:
一点编程也不会写的:零基础C语言学练课程
解决困扰你多年的C语言疑难杂症特性的C语言进阶课程
从零到写出一个爬虫的Python编程课程
只会语法写不出代码?手把手带你写100个编程真题的编程百练课程
信息学奥赛或C++选手的 必学C++课程
蓝桥杯ACM、信息学奥赛的必学课程:算法竞赛课入门课程
手把手讲解近五年真题的蓝桥杯辅导课程
发表评论 取消回复