无聊参加了一下北航的比赛，130多名吧 他们大一好多高手 我还是努力学习吧！
题目链接：http://acm.buaa.edu.cn/contest/63/problem/
作者解题报告：http://blog.csdn.net/dslovemz/article/details/8310089
题目 A：掷骰子
题目类型：简单概率
题目知识点 ：排序 简单分配
解题目过程：一看求期望值就觉得是个简单的题目，算了两个例子就知道是从小到大排，然后分配给每一个筛子。然后期望值酒是最小的了
代码如下
View Code
View Code题目 B：Taylor公式
题目类型：简单公式运用
题目知识点：浮点误差
解题过程：开始是用math.h里面的公式来的，然后测试一看，不是2012就是2011 然后就直接输出2012看看，没想到过了
附作者解题思路：
由tan(a+b) = (tan(a)+tan(b))/(1-tan(a)*tan(b))得
1/a = tan(arctan(1/b)+arctan(1/c)) = (1/b+1/c)/(1-1/b*1/c) = (b+c)/(cb-1)
=>cb-1=ab+ac => b*c-a*b-c*a=1
故ans = 2012/1 = 2012
代码如下：
View Code
1 #include <stdio.h>
2 #include <math.h>
3 int main()
4 {
5 int n,i;
6 scanf("%d",&n);
7 for (i = 1 ; i <= n ; i ++)
8 {
9 int a,b;
10 double c;
11 scanf("%d %d",&a,&b);
12
13 printf("2012\n");
14 }
15 return 0 ;
16
17 }
题目C：得了多少奖
题目类型：字符串分析
题目知识点：字符串查找 字符串处理 指针处理
解题过程 ：耗费了很多时间 发现在给出的页面边上就有一个得奖页面，将字符串按【分段 然后再用指针查找 最后打表 就可以直接输出了
代码如下：
View Code
1 #include <stdio.h>
2 #include <string.h>
3 struct st
4 {
5 char num[20];
6 int a,b,c,d;
7
8 };
9 int main()
10 {
11 struct st a[ ]={{"34060609",0,0,1,0},{"34060814",1,0,0,0},{"35060616",0,0,2,1},{"35060622",2,5,2,0},{"35060722",0,1,2,1},{"35090209",0,1,0,0},{"35211219",0,1,1,1},{"35211231",0,1,1,1},{"35211312",1,3,0,0},{"35211425",2,3,2,0},{"35230107",0,0,2,0},{"35230115",0,0,1,1},{"SY0906111",0,2,0,0},{"SY0906226",0,1,1,0},{"36030411",0,1,3,1},{"36060415",0,0,1,0},{"36060610",0,1,2,0},{"36090119",0,0,2,0},{"36091223",0,1,2,0},{"36093117",0,0,1,0},{"36211309",0,4,2,0},{"36211409",0,1,2,0},{"36211420",0,2,1,0},{"36230116",0,0,1,0},{"37060221",0,2,1,0},{"37071426",0,0,0,1},{"37211126",0,0,0,2},{"37211525",0,0,0,1},{"37211226",0,1,0,0},{"38022626",0,0,0,1},{"38211114",0,0,0,1},{"38211212",1,6,0,0},{"38211219",0,2,0,1},{"38211313",0,0,0,1},{"38211314",1,5,0,0},{"38211316",0,2,0,0},{"38211324",0,0,1,1},{"38211406",0,0,1,1},{"38211416",0,0,1,0},{"38211424",0,0,1,0},{"38211503",0,0,1,1},{"38211516",0,0,1,0},{"38230210",0,1,0,1},{"38230214",0,1,0,0},{"39055104",0,3,1,0},{"39061124",0,0,1,0},{"39061326",0,0,1,0},{"39061328",0,0,1,0},{"39061412",0,0,1,1},{"39061512",1,3,1,1},{"39211108",0,1,0,1},{"39211306",0,0,0,1},{"39211509",0,2,1,1},{"39231214",0,1,0,1},{"10061023",0,1,2,0},{"10061061",0,1,2,0},{"10061189",0,0,2,0},{"10061210",0,0,1,0},{"10091234",0,1,3,0},{"10131061",0,1,0,0},{"10131064",0,1,0,0},{"10211031",0,1,3,0},{"10211072",0,1,3,0},{"11061214",0,0,2,0},{"11061215",0,0,2,0},{"11211078",0,0,2,0},{"11211104",0,0,2,0},{"11211116",0,0,2,0},{"11211119",1,1,0,0}};
12 int i,j,k;
13 char c[20];
14 while (gets(c))
15 {
16 for (i = 0 ; i <= 68 ; i ++ )
17 if (strcmp(c,a[i].num) == 0)
18 printf("%dAu %dAg %dCu %dFe\n", a[i].a, a[i].b,a[i].c,a[i].d);
19
20 }
21 return 0;
22 }
题目D：GGCD
题目类型：数论
题目知识点：任何一个数都能用质数的幂相加而得到
题目描述：输入两个数 A B 求gcd（A，B^B）;
解题过程：开始一直以为是边乘边gcd然后判断，后来钦哥看了一下就给出解法了
解题思路：先将质数存数组，然后用另两个数组分别代表A 和 B 的每个质数因子的情况；然后將B的情况乘以B；然后他们公共的质数因子就是他们的最大公约数；
代码如下：
View Code
1 #include <stdio.h> 2 #include <math.h> 3 int min(int x, int y) 4 { 5 return x>=y?y:x; 6 } 7 int pw(int n,int t) 8 { 9 int i , s = 1; 10 for (i = 1 ; i <= t; i++ ) 11 s = s*n; 12 return s; 13 14 } 15 int main() 16 { 17 int a[11]={2,3,5,7,11,13,17,19,23,29,31}; 18 int x,y; 19 while (scanf("%d %d", &x, &y) != EOF) 20 { 21 int i , j , k = 1 ,t = y ; 22 int c[11]={0},b[11]={0}; 23 for(i = 0 ; i < 11 ; i ++) 24 { 25 if(x %a[i] == 0 ) 26 { c[i]++; 27 x = x / a[i]; 28 i = i - 1; 29 } 30 if (x == 1) 31 break; 32 } 33 for(i = 0 ; i < 11 ; i ++) 34 { 35 if(y % a[i] == 0 ) 36 { b[i]++; 37 y = y / a[i]; 38 i = i - 1; 39 } 40 if (y == 1) 41 break; 42 43 } 44 for (i = 0 ; i < 11; i ++) 45 b[i]=b[i]*t; 46 for (i =0 ; i < 11 ; i ++) 47 k = k*pw(a[i],min(c[i],b[i])); 48 49 printf("%d\n",k); 50 51 } 52 return 0; 53 54 55 }
其他题目没做
题目K：
和题目C一样 就不多说了；
代码如下：
View Code
1 #include <stdio.h> 2 #include <string.h> 3 char a[1000000]={0}; 4 int main() 5 { 6 int i=0 ; 7 char c, *p1, *p2; 8 memset(a ,0 ,sizeof(a)); 9 c = getchar(); 10 while (c != EOF) 11 { 12 a[i++] = c; 13 c = getchar (); 14 15 } 16 p1 = a; 17 while (strstr(p1,"Accepted")) 18 { 19 p1 = strstr(p1,"Accepted") - 628; 20 p2 = strstr(p1,"lang_"); 21 p2 = p2+5; 22 23 while (*p2 != '\'') 24 { 25 printf("%c", *p2); 26 p2 = p2+1; 27 } 28 printf(" "); 29 p2= strstr(p1,"/problem/"); 30 p2 = p2 + 9; 31 p1 = p2; 32 33 if (*p2 == '\"') 34 { 35 p2= strstr(p1,"/problem/"); 36 p2 = p2 + 9; 37 printf("%c ",*p2); 38 } 39 else 40 printf("%c ",*p2); 41 42 43 p2 = strstr(p1,"/profile/") + 9; 44 while (*p2 != '/') 45 { 46 printf("%c", *p2); 47 p2 = p2+1; 48 } 49 50 printf("\n"); 51 p1 = strstr(p1, "Accepted")+8 ; 52 } 53 54 return 0 ; 55 56 }