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杭电提交程序时后出现presentation error?该怎么解决

2012-03-08 
杭电提交程序时后出现presentation error?RT我做的是 1002 号题。A + B Problem IITime Limit: 2000/1000 M

杭电提交程序时后出现presentation error?
RT 
我做的是 1002 号题。
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 99748 Accepted Submission(s): 18931


Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

 

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

 

Sample Input
2
1 2
112233445566778899 998877665544332211
 

Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
 

Author
Ignatius.L
 我的代码没错误 输入输出也和他要求一样啊,怎么回事?

C/C++ code
#include <stdio.h>#include <math.h>#include <tchar.h>#include <string.h>#define  M 1001void AddResult(char *, char *);int main(int argc, TCHAR *argv[]){     int LineNum, i, j;    char StaFirst[M], StaSecond[M];    char SepFirst[20][M], SepSecond[20][M];                       scanf("%d", &LineNum);    getchar();    for (i = 0; i < LineNum; i++)    {        scanf("%s %s", StaFirst, StaSecond);        for (j = 0; StaFirst[j] != '\0'; j++)        {            SepFirst[i][j] = StaFirst[j];        }        SepFirst[i][j] = 0;        for (j = 0; StaSecond[j] != '\0'; j++)        {            SepSecond[i][j] = StaSecond[j];        }        SepSecond[i][j] = 0;    }    for (i = 0; i < LineNum; i++)    {                printf("Case %d:\n%s + %s = ", i + 1, SepFirst[i], SepSecond[i]);        AddResult(SepFirst[i], SepSecond[i]);        putchar(10);    }    getchar();    return 0;}void AddResult(char *SepFirst, char *SepSecond){    int Result[M] = { 0 }, Assistant[M] = { 0 }, LenFirst, LenSecond, Lenth;    int i, j, CarryFlag = 0;    LenFirst = strlen(SepFirst);    LenSecond = strlen(SepSecond);    if (LenFirst <= LenSecond)    {        for (i = LenFirst - 1, j = 0; i >= 0; i--, j++)        {            Assistant[j] = (int)(SepFirst[i] - '0');        }        for (i = LenSecond - 1, j = 0; i >= 0; i--, j++)        {            Result[j] = (Assistant[j] + (int)(SepSecond[i] - '0') + CarryFlag) % 10;            CarryFlag = (Assistant[j] + (int)(SepSecond[i] - '0') + CarryFlag) / 10;        }        if (CarryFlag == 1)        {            Result[j] = CarryFlag;            Lenth = j;        }        else            Lenth = j - 1;    }       else    {        for (i = LenSecond - 1, j = 0; i >= 0; i--, j++)        {            Assistant[j] = (int)(SepSecond[i] - '0');        }        for (i = LenFirst - 1, j = 0; i >= 0; i--, j++)        {            Result[j] = (Assistant[j] + (int)(SepFirst[i] - '0') + CarryFlag) % 10;            CarryFlag = (Assistant[j] + (int)(SepFirst[i] - '0') + CarryFlag) / 10;        }        if (CarryFlag == 1)        {            Result[j] = CarryFlag;            Lenth = j;        }        else            Lenth = j - 1;    }           for (i = Lenth; i >= 0; i--)    {        printf("%d", Result[i]);    }}





------解决方案--------------------


case1与case2要换行

[解决办法]
格式错误,请仔细检查case末的空行、case间的空行、每一行后的空格,等这些地方,看与题目描述是否一致。

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