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：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef struct{
    int pos, val;
}aa;

aa a[1010];

bool cmp(aa x,aa y)
{
    return x.val<y.val;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int T, n, k;
    scanf("%d", &T);
    for(int cnt = 1; cnt<=T; cnt++)
    {
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
        {
            a[i].pos = i;
            scanf("%d", &a[i].val);
        }
        scanf("%d", &k);
        sort(a+1, a+n+1, cmp);
        for(int i=1; i<=n; i++){
            if(a[i].pos == k){
                printf("%d
", i);
                break;
            }
        }

    }
    return 0;
}

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：

#include <iostream>
#include <cstdio>
#include <cstring>
#define MOD 1000000000+7
using namespace std;

long long a_b_MOD_c(long long a, long long b, long long c)
{
    if(b==1)
        return a%c;
    long long temp = a_b_MOD_c(a, b/2, c);
    if(b%2 == 1)
        return (temp*temp*a)%c;
    else
        return (temp*temp)%c;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int T;
    long long n;
    scanf("%d", &T);
    while(T--){
        scanf("%lld", &n);
        printf("%lld
", a_b_MOD_c(2,n,MOD)-1);
    }
    return 0;
}

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：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ll long long
using namespace std;

int tag[600000], sum[600000];

bool Judge(ll n)
{
    while(n){
        if(n%10==0 || n%10==1 || n%10==4 || n%10==9);
        else return false;
        n/=10;
    }
    return true;
}

int main()
{
//    freopen("in.txt", "r" , stdin);

    int T;
    ll a, b;
    memset(tag, 0, sizeof tag);
    memset(sum, 0, sizeof sum);
    for(ll i=1; i<100100; i++){
        if(Judge(i*i)){
            tag[i] = 1;
        }
        sum[i] = sum[i-1]+tag[i];
    }
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld%lld", &a, &b);
        int temp = (sqrt(a)-(int)sqrt(a))==0?1:0;
        int ans = sum[(int)sqrt(b)] - sum[(int)sqrt(a)-temp];
        printf("%d
", ans);
    }
    return 0;
}

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：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll unsigned long long
#define MOD 1000000000+7
using namespace std;

ll a_b_MOD_c(ll a, ll b, ll c)
{
    ll product = a,ans = 1;
    while(b)
    {
        if(b&1==1){
            ans *= product;
            ans %= c;
        }
        b>>=1;
        product = (product*product)%c;
    }
    return ans;
}

ll Change(ll n)
{
    int dig[1000], top = 0;
    while(n){
        dig[top++] = (n&1)?1:0;
        n>>=1;
    }
    ll ret = 0;
    for(int i=top-1; i>=0; i--){
        ret += dig[i]==1?1:0;
        if(i!=0)ret*=10;
    }
    return ret;
}

int main()
{
//    freopen("in.txt", "r" , stdin);

    int T;
    ll n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%llu", &n);
        ll ans = a_b_MOD_c(2, Change(n), MOD);
        ans = a_b_MOD_c(ans, 2, MOD);
        printf("%llu
", ans);
    }
    return 0;
}