三题，B、C两题出太晚了，，尤其是B题，想复杂了，其实想到就很简单了哎。。有时候思路要放水一点。
D题很恶心不考虑了。E题是树状数组加序DFS，。果断不会，该学习下树状数组了。。
Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (x,?y), he can move to (or attack) positions (x?+?1,?y), (x–1,?y), (x,?y?+?1) and (x,?y–1).
Iahub wants to know how many Coders can be placed on an n?×?n chessboard, so that no Coder attacks any other Coder.
The first line contains an integer n (1?≤?n?≤?1000).
On the first line print an integer, the maximum number of Coders that can be placed on the chessboard.
On each of the next n lines print n characters, describing the configuration of the Coders. For an empty cell print an ‘.‘, and for a Coder print a ‘C‘.
If there are multiple correct answers, you can print any.
2
2 C. .C
A：给一个棋盘，放卒子，最多放几个和方法。
隔开放就可以了水题。
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n;
int main() {
scanf("%d", &n);
printf("%d\n", (n * n + 1) / 2);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if ((i + j) % 2 == 0) printf("C");
else printf(".");
}
printf("\n");
}
return 0;
}
Iahub wants to enhance his multitasking abilities. In order to do this, he wants to sort n arrays simultaneously, each array consisting of mintegers.
Iahub can choose a pair of distinct indices i and j (1?≤?i,?j?≤?m,?i?≠?j). Then in each array the values at positions i and j are swapped only if the value at position i is strictly greater than the value at position j.
Iahub wants to find an array of pairs of distinct indices that, chosen in order, sort all of the n arrays in ascending or descending order (the particular
order is given in input). The size of the array can be at most (at
most
pairs).
Help Iahub, find any suitable array.
The first line contains three integers n (1?≤??n?≤?1000), m (1?≤?m?≤??100) and k. Integer k is 0 if the arrays must be sorted in ascending order, and 1 if the arrays must be sorted in descending order. Each line i of the next n lines contains m integers separated by a space, representing the i-th array. For each element x of the array i, 1?≤?x?≤?106 holds.
On the first line of the output print an integer p, the size of the array (p can
be at most ).
Each of the next p lines must contain two distinct integers i and j (1?≤?i,?j?≤?m,?i?≠?j),
representing the chosen indices.
If there are multiple correct answers, you can print any.
2 5 0 1 3 2 5 4 1 4 3 2 5
3 2 4 2 3 4 5
3 2 1 1 2 2 3 3 4
1 2 1
思路：一开始以为很复杂的。后来想到其实只要按最多的步数|1 2 2 3 3 4 4 5 | 1 2 2 3 3 4 | 1 2 2 3 |1 2|类似这样就执行就可以了。
代码：
#include <stdio.h>
#include <string.h>
int n, m, k;
int main () {
int i, j;
scanf("%d%d%d", &n, &m, &k);
for (i = 1; i <= n; i++)
for (j = 1; j <= m; j++)
scanf("%*d");
printf("%d\n", m * (m - 1) / 2);
if (k == 0) {
for (i = m - 1; i >= 1; i--)
for (j = 1; j <= i; j++) {
printf("%d %d\n", j, j + 1);
}
}
else {
for (i = m; i > 1; i--)
for (j = m; j > 1 + m - i; j--) {
printf("%d %d\n", j, j - 1);
}
}
return 0;
}
Iahub helps his grandfather at the farm. Today he must milk the cows. There are n cows sitting in a row, numbered from 1 to n from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
The first line contains an integer n (1?≤?n?≤?200000). The second line contains n integers a1, a2, ..., an, where ai is 0 if the cow number i is facing left, and 1 if it is facing right.
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
4 0 0 1 0
1
5 1 0 1 0 1
3
思路：先把朝右的按从左往右挤掉，在把朝左的从右往左挤掉，这是最优的。。仔细想想就明白。中间一步用前缀和保存惊吓的牛数。
代码：
#include <stdio.h>
#include <string.h>
int n;
__int64 num[200005], sum[200005];
int main() {
scanf("%d", &n);
int i;
for (i = 1; i <= n; i++) {
scanf("%I64d", &num[i]);
}
num[n + 1] = -1;
for (i = n + 1; i >= 1; i--)
if (num[i] == 0) {
sum[i - 1] = sum[i] + 1;
}
else sum[i - 1] = sum[i];
__int64 ans = 0;
for (i = 1; i <= n; i++)
if (num[i] == 1) {
ans += sum[i];
}
printf("%I64d\n", ans);
return 0;
}