题目：
Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.
Input
Input starts with an integer T (≤ 525), denoting the number of test cases.
Each case starts with a line containing two integers a (-10200 ≤ a ≤ 10200) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.
Output
For each case, print the case number first. Then print 'divisible' if a is divisible by b. Otherwise print 'not divisible'.
Sample Input
6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
Sample Output
Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
题意描述：
大整数取模。
解题思路：
1、模拟大整数除法。2、从低位到高位每次乘以位权对除数取模。
AC代码：
比赛时候写的比较仓促，使用了第一种较笨的方法。
1 #include<stdio.h> 2 #include<string.h> 3 int f(char s1[],char s2[]); 4 int sub(int *p1,int *p2,int l1,int l2); 5 char s1[300],s2[300]; 6 int an1[300],an2[300]; 7 int main() 8 { 9 int T,flag,t=1,l1,l2; 10 scanf("%d",&T); 11 while(T--) 12 { 13 memset(s1,0,sizeof(s1));//注意重置字符数组 14 memset(s2,0,sizeof(s2));//注意重置字符数组 15 scanf("%s%s",s1,s2); 16 flag=0; 17 if(s1[0]=='-' || s2[0]=='-') 18 { 19 if(s1[0]=='-'&&s2[0]=='-') 20 flag=f(s1+1,s2+1); 21 else 22 { 23 if(s1[0]=='-') 24 flag=f(s1+1,s2); 25 else 26 flag=f(s1,s2+1); 27 } 28 } 29 else 30 flag=f(s1,s2); 31 32 //printf("%d ",flag); 33 if(flag)//1能整除，0不能整除 34 printf("Case %d: divisible ",t++); 35 else 36 printf("Case %d: not divisible ",t++); 37 } 38 } 39 40 int f(char s1[],char s2[]) 41 { 42 int l1,l2,b,i,j; 43 l1=strlen(s1); 44 l2=strlen(s2); 45 if(l1==1&&s1[0]=='0') 46 return 1; 47 48 if(l1<l2) 49 return 0; 50 memset(an1,0,sizeof(an1)); 51 memset(an2,0,sizeof(an2)); 52 j=0; 53 for(i=l1-1;i>=0;i--) 54 an1[j++]=s1[i]-'0'; 55 j=0; 56 for(i=l2-1;i>=0;i--) 57 an2[j++]=s2[i]-'0'; 58 59 l1=sub(an1,an2,l1,l2); 60 if(l1 < 0) 61 return 0; 62 else if(l1==0) 63 return 1; 64 65 int nt=l1-l2; 66 if(nt < 0) 67 return 0; 68 else if(nt > 0) 69 { 70 for(i=l1-1;i>=0;i--) 71 { 72 if(i >= nt) 73 an2[i]=an2[i-nt]; 74 else 75 an2[i]=0; 76 } 77 } 78 l2=l1; 79 for(j=0;j<=nt;j++) 80 { 81 int tem; 82 while( (tem=sub(an1,an2+j,l1,l2-j)) >= 0) 83 { 84 l1=tem; 85 } 86 } 87 88 if(l1==0) 89 return 1; 90 else 91 return 0; 92 } 93 int sub(int *p1,int *p2,int l1,int l2) 94 { 95 96 int i; 97 if(l1 < l2) 98 return -1; 99 100 bool bl=false; 101 if(l1 == l2) 102 { 103 for(i=l1-1;i>=0;i--) 104 { 105 if(p1[i] > p2[i]) 106 bl=true; 107 else 108 if(p1[i] < p2[i]) 109 { 110 if(!bl) 111 return -1; 112 } 113 } 114 } 115 for(i=0;i<l1;i++) 116 { 117 p1[i] -= p2[i]; 118 if(p1[i] < 0) 119 { 120 p1[i] += 10; 121 p1[i + 1]--; 122 } 123 } 124 for(i=l1-1;i>=0;i--) 125 { 126 if(p1[i]) 127 return i+1; 128 } 129 return 0; 130 }