A.​​Ilya and Bank Account​​

Ilya得到了一个礼物，可以在删掉银行账户最后和倒数第二位的数字（账户有可能是负的），也可以不做任何处理。

//codeforces 313A 
//2013-05-31-13.47
#include <stdio.h>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    scanf("%d", &n);
    if (n)
    {
        if (n >= 0)
        {
            printf("%d\n", n);
        }
        else
        {
            n = -n;
            int a = n/10;
            int b = n/100 * 10 + n%10;
            printf("%d\n", -min(a, b));
        }
    }
    return 0;
}


B.​​Ilya and Queries​​

给你一个字符串，然后有M个询问，寻问的是从l到r之间有多少对字符串满足 s[i] == s[i+1]。

简单的树状数组题目

//codeforces 313 B
//2013-05-31-14.15
#include <stdio.h>
#include <string.h>
const int maxn = 100005;
char s[maxn];
int a[maxn];
int n;
inline int lowbit(int x)
{
    return x&-x;
}

void update(int x)
{
    while (x <= n)
    {
        a[x] += 1;
        x += lowbit(x);
    }
}

int getsum(int x)
{
    int sum = 0;
    while (x)
    {
        sum += a[x];
        x -= lowbit(x);
    }
    return sum;
}

int main()
{
    while (scanf("%s", s) != EOF)
    {
        int m, l, r;
        n = strlen(s);
        memset(a, 0, sizeof(a));
        for (int i = 0; i < n-1; i++)
        {
            if (s[i] == s[i+1])
                update(i+1);
        }
        scanf("%d", &m);
        while (m--)
        {
            scanf("%d %d", &l, &r);
            printf("%d\n", getsum(r-1) - getsum(l-1));
        }
    }
    return 0;
}

C.​​Ilya and Matrix​​

给你4^n个数，让你放进那个2^n * 2^n的矩阵里让这个矩阵beauty值最大，beauty值的计算方法题里说了。。。

The beauty of a 2n × 2n-sized matrix is an integer, obtained by the following algorithm:


Find the maximum element in the matrix. Let's denote it as m.
If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2n - 1 × 2n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.

As you can see, the algorithm is recursive

贪心吧，贪心就行，排个序解决了

//codeforces 313c
//2013-06-03-15.55
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 2*1000006;
__int64 a[maxn];

bool cmp(int x, int y)
{
    return x > y;
}

int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; i++)
            scanf("%I64d", &a[i]);
        sort(a+1, a+n+1, cmp);
        int x = n;
        __int64 ans = 0;
        while (x)
        {
            for (int i = 1; i <= x; i++)
                ans += a[i];
            x /= 4;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}


D.​​Ilya and Roads​​


E.​​Ilya and Two Numbers​​