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;
}
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;
}
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;
}
