UVA465 Overflow
高精度运算，涉及加法、乘法和比大小。
一不留神就写了百行，WA的原因是output没有把输入再打印一遍。不过改的时候也考虑了一下前导零，用了clean()函数。可能不考虑也能A，题里面又没说嘛。
题目：
Overflow
| Overflow |
Write a program that reads an expression consisting of two non-negative integer and an operator. Determine if either integer or the result of the expression is too large to be represented as a ``normal'' signed integer (type integer if you are working Pascal, type int if you are working in C).
Input
An unspecified number of lines. Each line will contain an integer, one of the two operators + or *, and another integer.
Output
For each line of input, print the input followed by 0-3 lines containing as many of these three messages as are appropriate: ``first number too big'', ``second number too big'', ``result too big''.
Sample Input
300 + 3 9999999999999999999999 + 11
Sample Output
300 + 3 9999999999999999999999 + 11 first number too big result too big
解答：
1 #include<stdio.h> 2 #include<string.h> 3 #include<limits.h> 4 #include<stdlib.h> 5 #include<ctype.h> 6 struct bign 7 { 8 int len; 9 int s[500]; 10 }; 11 char str[1010],it[3][50]; 12 bign a,b,res,iMax; 13 int max(int a,int b) 14 { 15 return (a>b)?a:b; 16 } 17 void clean(bign *b) 18 { 19 while(b->len&&b->s[b->len-1]==0) 20 b->len--; 21 } 22 void change(char *str,bign *b) 23 { 24 int i; 25 b->len=strlen(str); 26 for(i=0;i<b->len;i++) 27 b->s[i]=str[b->len-i-1]-'0'; 28 clean(b); 29 } 30 void doit() 31 { 32 char tmp[500]; 33 sprintf(it[0],"%s","first number too big"); 34 sprintf(it[1],"%s","second number too big"); 35 sprintf(it[2],"%s","result too big"); 36 sprintf(tmp,"%d",INT_MAX); 37 change(tmp,&iMax); 38 } 39 void mkstr() 40 { 41 char tmp[500]; 42 int i,p=0; 43 for(i=0;isdigit(str[i]);i++) 44 tmp[i]=str[i]; 45 tmp[i]='\0'; 46 change(tmp,&a); 47 for(i=i+3;isdigit(str[i]);i++) 48 { 49 tmp[p]=str[i]; 50 p++; 51 } 52 tmp[p]='\0'; 53 change(tmp,&b); 54 } 55 void plus(bign *a,bign *b,bign *res) 56 { 57 int i,g; 58 for(i=0,g=0;g!=0||i<max(a->len,b->len);i++) 59 { 60 int x=g; 61 if(i<a->len) 62 x+=a->s[i]; 63 if(i<b->len) 64 x+=b->s[i]; 65 res->s[i]=x%10; 66 g=x/10; 67 } 68 res->len=i; 69 clean(res); 70 } 71 void mult(bign *a,bign *b,bign *res) 72 { 73 int i,j; 74 res->len=a->len+b->len; 75 for(i=0;i<a->len;i++) 76 { 77 for(j=0;j<b->len;j++) 78 { 79 res->s[i+j]+=a->s[i]*b->s[j]; 80 } 81 } 82 for(i=0;i<res->len-1;i++) 83 { 84 res->s[i+1]+=res->s[i]/10; 85 res->s[i]=res->s[i]%10; 86 } 87 clean(res); 88 } 89 int bigncmp(bign *a,bign *b) 90 { 91 if(a->len!=b->len) 92 { 93 if(a->len>b->len) 94 return 1; 95 else 96 return -1; 97 } 98 int i; 99 for(i=a->len-1;i>=0;i--) 100 { 101 if(a->s[i]!=b->s[i]) 102 { 103 if(a->s[i]>b->s[i]) 104 return 1; 105 else 106 return -1; 107 } 108 } 109 return 0; 110 } 111 int main() 112 { 113 doit(); 114 while(gets(str)!=NULL) 115 { 116 puts(str); 117 int i; 118 memset(res.s,0,sizeof(res.s)); 119 mkstr(); 120 if(bigncmp(&a,&iMax)>0) 121 { 122 puts(it[0]); 123 } 124 if(bigncmp(&b,&iMax)>0) 125 { 126 puts(it[1]); 127 } 128 if(strstr(str,"*")!=NULL) 129 { 130 mult(&a,&b,&res); 131 if(bigncmp(&res,&iMax)>0) 132 puts(it[2]); 133 } 134 else 135 { 136 plus(&a,&b,&res); 137 if(bigncmp(&res,&iMax)>0) 138 puts(it[2]); 139 } 140 } 141 return 0; 142 }
