UVA 465 Overflow
2015-05-15 20:25
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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'' signedinteger (type integer if you are working Pascal, type intif 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 toobig''.
Sample Input
300 + 3 9999999999999999999999 + 11
Sample Output
300 + 3 9999999999999999999999 + 11
first number too big
result too big
题意:
给一个表达式,判断这两个数与表达式的结果是否有大于 int 最大值的,
刚开始理解错了题意,WA好多次!double 保存结果,不超。
代码:
#include <cstdio> #include <cstring> #include<climits> // INT_MAX 需要 int main() { char s1[1000], s2[1000], ch; int la, lb; while( ~scanf("%s %c %s", &s1, &ch, &s2) ) { double sum1 = 0, sum2 = 0; la = strlen(s1); lb = strlen(s2); for(int i = 0; i < la; i++) sum1 = 10 * sum1 + s1[i] - '0'; for(int i = 0; i < lb; i++) sum2 = 10 * sum2 + s2[i] - '0'; printf("%s %c %s\n", s1, ch, s2); if(sum1 > INT_MAX) printf("first number too big\n"); if(sum2 > INT_MAX) printf("second number too big\n"); if(ch == '+') sum1 += sum2 ; if(ch == '*') sum1 *= sum2 ; if(sum1 > INT_MAX) printf("result too big\n"); } return 0; }
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