GreatFeng的专栏

#include<stdio.h> #include<string.h> int main(){ int n,a[1001],i,j,k,m,max; char str[1001][16],str1[1001][16]; while(scanf("%d",&n)!=EOF,n) { getchar();

1002 A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 69615    Accepted Submission(s): 12678 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 #include <stdio.h> #include <string.h> int main(){ char str1[1001], str2[1001]; int t, i, len_str1, len_str2, len_max, num = 1, k; scanf("%d", &t); getchar(); while(t--){ int a[1001] = {0}, b[1001] = {0}, c[1001] = {0}; scanf("%s", str1); len_str1 = strlen(str1); for(i = 0; i <= len_str1 - 1; ++i) a[i] = str1[len_str1 - 1 - i] - '0'; scanf("%s",str2); len_str2 = strlen(str2); for(i = 0; i <= len_str2 - 1; ++i) b[i] = str2[len_str2 - 1 - i] - '0'; if(len_str1 > len_str2) len_max = len_str1; else len_max = len_str2; k = 0; for(i = 0; i <= len_max - 1; ++i){ c[i] = (a[i] + b[i] + k) % 10; k = (a[i] + b[i] + k) / 10; } if(k != 0) c[len_max] = 1; printf("Case %d:\n", num); num++; printf("%s + %s = ", str1, str2); if(c[len_max] == 1) printf("1"); for(i = len_max - 1; i >= 0; --i){ printf("%d", c[i]); } printf("\n"); if(t >= 1) printf("\n"); } return 0; }

二维数组所有考点总结 2010-10-21 所有用到的的数组是：a[3][3]={{23,46,11},{99,45,82},{72,90,21}}; 一：遍历打印输出二维数组的元素 #include <stdio.h> void main(){ int a[3][3]={{23,46,11},{99,45,82},{72,90,21}}; for(int i=0;i<3;i++){ for(int j=0;j<3;j++){ printf("%d ",a[i][j]); } printf("\n"); } } 思路：【1：每一个下标号对应的元素值是唯一的； 2：因为是二维数组，必须要遍历行，列所以需要两个循环变量； 3：在每一次循环遍历列数后，就用换行语句，可以打印输出标准的行列格式】

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