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为题如下： 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 int
#include<stdio.h> #include<stdlib.h> typedef int item; typedef struct node { item data; &
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2920 Accepted Submission(s): 19
实现代码： 单链表 /****************************************************** > File Name: Link.c > Author: wangqiang &
A + B Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8165&
#include<stdio.h> #include<stdlib.h> #include<limits.h> int** buildTsble(void); void fillTable(int** table); void processTable(int** table); int smaller(int first
#include<stdio.h> #include<string.h> int main() { long long z,i,x=0,y,b,n; while
寻找大富翁 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1164 Accepted Submission(s): 597
xxx定律 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 750 Accepted Submission(s): 621 P
shǎ崽 OrOrOrOrz Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3630 Accepted Submission(s)
Saving HDU Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2834 Accepted Submission(s): 12
Who's in the Middle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4519 Accepted Submissi
简易版之最短距离 时间限制(普通/Java):1000MS/10000MS 运行内存限制:65536KByte 总提交: 357 测试通过: 160 描述 寒假的时候，ACBOY要去拜访很多朋友，恰巧他所有朋友的家都处在坐标平面的X轴
超级简单的数学题 有5人坐在一起，当问第5个人多少岁,他说比第4个人大2岁，问第4个人多少岁,他说比第3个人大2岁，依此下去，问第一个人多少岁，他说他10岁，最后求第5个人多少岁 如果所坐的不是5人而是n人，写出第n个人的年龄表达式。 化简后的公式: F(n)=10+(n-1)*2
欧几里德算法 最大公约数（greatest common divisor，简写为gcd；或highest common factor，简写为hcf)，指某几个整数共有因子中最大的一个。 最小公倍数（Least Common Multiple，缩写L.C.M.），lcm(m,n) = m * n / gcd(m,n)。有时，由于数字较大，m*n可能过界，所以最好写成 m / gc
将Julian历法中的日转换成月和日 #include<stdio.h> int a[15]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int i,x; int fun(int *p,int x) { if(x>=366||x<1) { printf(&
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