CodeChef November Challenge 2013 解题报告

Uncle Johny

题目链接：http://www.codechef.com/NOV13/problems/JOHNY

水题。排序即可

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 typedef struct{
 8     int pos, val;
 9 }aa;
10 
11 aa a[1010];
12 
13 bool cmp(aa x,aa y)
14 {
15     return x.val<y.val;
16 }
17 
18 int main()
19 {
20 //    freopen("in.txt", "r", stdin);
21 
22     int T, n, k;
23     scanf("%d", &T);
24     for(int cnt = 1; cnt<=T; cnt++)
25     {
26         scanf("%d", &n);
27         for(int i=1; i<=n; i++)
28         {
29             a[i].pos = i;
30             scanf("%d", &a[i].val);
31         }
32         scanf("%d", &k);
33         sort(a+1, a+n+1, cmp);
34         for(int i=1; i<=n; i++){
35             if(a[i].pos == k){
36                 printf("%d\n", i);
37                 break;
38             }
39         }
40 
41     }
42     return 0;
43 }
View Code

Missing some chairs

题目链接：http://www.codechef.com/NOV13/problems/MCHAIRS

计算2^n - 1。快速模幂

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #define MOD 1000000000+7
 5 using namespace std;
 6 
 7 long long a_b_MOD_c(long long a, long long b, long long c)
 8 {
 9     if(b==1)
10         return a%c;
11     long long temp = a_b_MOD_c(a, b/2, c);
12     if(b%2 == 1)
13         return (temp*temp*a)%c;
14     else
15         return (temp*temp)%c;
16 }
17 
18 int main()
19 {
20 //    freopen("in.txt", "r", stdin);
21 
22     int T;
23     long long n;
24     scanf("%d", &T);
25     while(T--){
26         scanf("%lld", &n);
27         printf("%lld\n", a_b_MOD_c(2,n,MOD)-1);
28     }
29     return 0;
30 }
View Code

Square Digit Squares

题目链接：http://www.codechef.com/NOV13/problems/SDSQUARE

找到1-1e10之间的平方数并且只含有0.1.4.9

暴力枚举平方数，打表即可

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #define ll long long
 7 using namespace std;
 8 
 9 int tag[600000], sum[600000];
10 
11 bool Judge(ll n)
12 {
13     while(n){
14         if(n%10==0 || n%10==1 || n%10==4 || n%10==9);
15         else return false;
16         n/=10;
17     }
18     return true;
19 }
20 
21 int main()
22 {
23 //    freopen("in.txt", "r" , stdin);
24 
25     int T;
26     ll a, b;
27     memset(tag, 0, sizeof tag);
28     memset(sum, 0, sizeof sum);
29     for(ll i=1; i<100100; i++){
30         if(Judge(i*i)){
31             tag[i] = 1;
32         }
33         sum[i] = sum[i-1]+tag[i];
34     }
35     scanf("%d", &T);
36     while(T--)
37     {
38         scanf("%lld%lld", &a, &b);
39         int temp = (sqrt(a)-(int)sqrt(a))==0?1:0;
40         int ans = sum[(int)sqrt(b)] - sum[(int)sqrt(a)-temp];
41         printf("%d\n", ans);
42     }
43     return 0;
44 }
View Code

Superpowers of 2

题目链接：http://www.codechef.com/NOV13/problems/SPOTWO

也是快速模幂，数据卡的比较死一定要用unsigned long long才行，最后平方一定要分开做

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #define ll unsigned long long
 6 #define MOD 1000000000+7
 7 using namespace std;
 8 
 9 ll a_b_MOD_c(ll a, ll b, ll c)
10 {
11     ll product = a,ans = 1;
12     while(b)
13     {
14         if(b&1==1){
15             ans *= product;
16             ans %= c;
17         }
18         b>>=1;
19         product = (product*product)%c;
20     }
21     return ans;
22 }
23 
24 ll Change(ll n)
25 {
26     int dig[1000], top = 0;
27     while(n){
28         dig[top++] = (n&1)?1:0;
29         n>>=1;
30     }
31     ll ret = 0;
32     for(int i=top-1; i>=0; i--){
33         ret += dig[i]==1?1:0;
34         if(i!=0)ret*=10;
35     }
36     return ret;
37 }
38 
39 int main()
40 {
41 //    freopen("in.txt", "r" , stdin);
42 
43     int T;
44     ll n;
45     scanf("%d", &T);
46     while(T--)
47     {
48         scanf("%llu", &n);
49         ll ans = a_b_MOD_c(2, Change(n), MOD);
50         ans = a_b_MOD_c(ans, 2, MOD);
51         printf("%llu\n", ans);
52     }
53     return 0;
54 }
View Code

 

posted @ 2013-11-01 23:55  张小豪  阅读(429)  评论(0)  编辑  收藏  举报