HDU 4923 Room and Moor
题意：给出A序列 求满足题目所写的B序列 使得方差最小
思路：可以想到最后的结果中 B序列的值一定是一段一段的 那么我们可以类似贪心去搞 对于一段序列我们可以求出什么样的b值使得方差最小 即序列中1的个数除以序列长度 又因为B是单调的 可以用一个单调栈去模拟 复杂度远远小于n^2 不要被吓怕…
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010
int t, n, top;
int a[N], f[2][N];
struct st {
int head;
double val;
} g[N];
int main() {
int i, idx, num;
double tmp1, ans;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (i = 1; i <= n; i++) {
scanf("%d", &a[i]);
f[0][i] = f[0][i - 1];
f[1][i] = f[1][i - 1];
f[a[i]][i]++;
}
f[0][n + 1] = f[0][n];
f[1][n + 1] = f[1][n];
g[0].head = 1;
g[0].val = -1;
top = 0;
for (i = 1; i <= n; i++) {
idx = i;
tmp1 = a[i];
while (top >= 0 && g[top].val > tmp1) {
idx = g[top].head;
tmp1 = 1.0 * (f[1][i] - f[1][idx - 1]) / (i - idx + 1);
top--;
}
top++;
g[top].head = idx;
g[top].val = tmp1;
}
ans = 0;
g[top + 1].head = n + 1;
for (i = 1; i <= top; i++) {
tmp1 = g[i].val;
num = f[0][g[i + 1].head - 1] - f[0][g[i].head - 1];
ans += tmp1 * tmp1 * num;
num = f[1][g[i + 1].head - 1] - f[1][g[i].head - 1];
ans += (1.0 - tmp1) * (1.0 - tmp1) * num;
}
printf("%.6f\n", ans);
}
return 0;
}HDU 4925 Apple Tree
题意：n*m的格子 要么种苹果 要么施化肥 施肥后的格子的相邻格子如果种了苹果 则苹果数翻倍 问最多产生几个苹果
思路：明显就是按矩形黑白染色的方法来种树和施肥 那么枚举两种情况 即是黑格子还是白格子种苹果 最后取max
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int t, n, m;
int main() {
int i, j, ans, res, k;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
res = ans = 0;
for (i = 1; i <= n; i++) {
for (j = 1; j <= m; j++) {
k = 1;
if (i + 1 <= n)
k *= 2;
if (i - 1 >= 1)
k *= 2;
if (j + 1 <= m)
k *= 2;
if (j - 1 >= 1)
k *= 2;
if ((i + j) & 1)
ans += k;
else
res += k;
}
}
printf("%d\n", max(ans, res));
}
return 0;
}HDU 4927 Series1
题意：一个序列A 每次根据这个方程ai=ai+1-ai使整个序列数字个数-1 问n-1次操作后剩下哪个数
思路：水题 只要把式子化简即可 注意高精度
代码：
import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int t, n, T, i, j;
BigInteger[] a = new BigInteger[3010];
BigInteger[] c = new BigInteger[3010];
T = cin.nextInt();
for (t = 1; t <= T; t++) {
n = cin.nextInt();
c[0] = BigInteger.ONE;
for (i = 1; i <= n; i++)
c[i] = c[i - 1].multiply(BigInteger.valueOf(n - i)).divide(
BigInteger.valueOf(i));
for (i = 1; i <= n; i++)
a[i] = cin.nextBigInteger();
BigInteger ans;
ans = BigInteger.ZERO;
int sign = 1;
for (j = n; j >= 1; j--) {
if (sign == 1) {
ans = ans.add(a[j].multiply(c[n - j]));
sign = -1;
} else {
ans = ans.subtract(a[j].multiply(c[n - j]));
sign = 1;
}
}
System.out.println(ans);
}
cin.close();
}
}HDU 4930 Fighting the Landlords
题意：斗地主- -b 如果一下把手牌出光就算赢 或者 你出一次对手管不上就算赢 问能不能赢
思路：模拟题… 细心就好 逻辑如下
先看是否一次能出完 如果不能 判断大小王 在自己手里必胜在别人手里必输
如果还没有结果 判炸弹 如果还没结果 就分所有能出牌的可能判断
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int kind[10][5];
int cas, flag;
char str[20];
int p1[20], p2[20], vis[20], only[10][5];
int change(char x) {
if (x == 'Y')
return 17;
else if (x == 'X')
return 16;
else if (x == '2')
return 15;
else if (x == 'A')
return 14;
else if (x == 'K')
return 13;
else if (x == 'Q')
return 12;
else if (x == 'J')
return 11;
else if (x == 'T')
return 10;
else
return x - '0';
}
bool cmp(int a, int b) {
return a > b;
}
int main() {
int i, j, len1, len2;
scanf("%d", &cas);
getchar();
while (cas--) {
memset(kind, -1, sizeof(kind));
memset(only, 0, sizeof(only));
flag = 0;
gets(str);
len1 = strlen(str);
for (i = 0; i < len1; i++) {
p1[i] = change(str[i]);
}
gets(str);
len2 = strlen(str);
for (i = 0; i < len2; i++) {
p2[i] = change(str[i]);
}
sort(p1, p1 + len1, cmp);
sort(p2, p2 + len2, cmp);
//7
if (len1 >= 2 && p1[0] == 17 && p1[1] == 16) {
kind[7][1] = 1;
// printf("1\n");
printf("Yes\n");
continue;
}
if (len2 >= 2 && p2[0] == 17 && p2[1] == 16) {
kind[7][2] = 1;
if (len2 == 2)
only[7][2] = 1;
}
//8
for (i = 3; i < len1; i++) {
if (p1[i - 3] == p1[i - 2] && p1[i - 2] == p1[i - 1]
&& p1[i - 1] == p1[i]) {
kind[8][1] = p1[i];
if (len1 == 4) {
// printf("2\n");
printf("Yes\n");
flag = 1;
}
break;
}
}
if (flag)
continue;
for (i = 3; i < len2; i++) {
if (p2[i - 3] == p2[i - 2] && p2[i - 2] == p2[i - 1]
&& p2[i - 1] == p2[i]) {
kind[8][2] = p2[i];
if (len2 == 4)
only[8][2] = 1;
break;
}
}
if ((kind[8][1] > kind[8][2]) && (kind[7][2] == -1)) {
// printf("3\n");
printf("Yes\n");
continue;
}
//1
kind[1][1] = p1[0];
kind[1][2] = p2[0];
if (len1 == 1) {
// printf("4\n");
printf("Yes\n");
continue;
}
if (kind[1][1] != -1 && kind[1][1] >= kind[1][2]) {
if ((kind[7][2] == -1) && (kind[8][2] == -1)) {
// printf("5\n");
printf("Yes\n");
continue;
}
}
//2
for (i = 1; i < len1; i++) {
if (p1[i - 1] == p1[i]) {
kind[2][1] = p1[i];
if (len1 == 2)
only[2][1] = 1;
break;
}
}
if (only[2][1]) {
//printf("6\n");
printf("Yes\n");
continue;
}
for (i = 1; i < len2; i++) {
if (p2[i - 1] == p2[i]) {
kind[2][2] = p2[i];
break;
}
}
//printf("2....%d %d\n",kind[2][1],kind[2][2]);
if (kind[2][1] != -1 && kind[2][1] >= kind[2][2]) {
if ((kind[7][2] == -1) && (kind[8][2] == -1)) {
//printf("7\n");
printf("Yes\n");
continue;
}
}
//3
for (i = 2; i < len1; i++) {
if (p1[i - 1] == p1[i] && p1[i - 2] == p1[i - 1]) {
kind[3][1] = p1[i];
if (len1 == 3)
only[3][1] = 1;
break;
}
}
if (only[3][1]) {
//printf("8\n");
printf("Yes\n");
continue;
}
for (i = 2; i < len2; i++) {
if (p2[i - 1] == p2[i] && p2[i - 2] == p2[i - 1]) {
kind[3][2] = p2[i - 1];
break;
}
}
if (kind[3][1] != -1 && kind[3][1] >= kind[3][2]) {
if ((kind[7][2] == -1) && (kind[8][2] == -1)) {
//printf("9\n");
printf("Yes\n");
continue;
}
}
//4
for (i = 2; i < len1; i++) {
if (p1[i - 1] == p1[i] && p1[i - 2] == p1[i - 1]) {
if ((i + 1 < len1 && p1[i + 1] != p1[i])
|| (i + 2 < len1 && p1[i + 2] != p1[i])
|| (i - 3 >= 0 && p1[i - 3] != p1[i])
|| (i - 4 >= 0 && p1[i - 4] != p1[i])) {
kind[4][1] = p1[i];
if (len1 == 4)
only[4][1] = 1;
break;
}
}
}
if (only[4][1]) {
printf("Yes\n");
continue;
}
for (i = 2; i < len2; i++) {
if (p2[i - 1] == p2[i] && p2[i - 2] == p2[i - 1]) {
if ((i + 1 < len2 && p2[i + 1] != p2[i])
|| (i + 2 < len2 && p2[i + 2] != p2[i])
|| (i - 3 >= 0 && p2[i - 3] != p2[i])
|| (i - 4 >= 0 && p2[i - 4] != p2[i])) {
kind[4][2] = p2[i];
break;
}
}
}
if (kind[4][1] != -1 && kind[4][1] >= kind[4][2]) {
if ((kind[7][2] == -1) && (kind[8][2] == -1)) {
printf("Yes\n");
continue;
}
}
//5
memset(vis, 0, sizeof(vis));
for (i = 2; i < len1; i++) {
if (p1[i - 1] == p1[i] && p1[i - 2] == p1[i - 1]) {
vis[i - 2] = 1;
vis[i - 1] = 1;
vis[i] = 1;
for (j = 1; j < len1; j++) {
if ((!vis[j - 1]) && (!vis[j]) && p1[j] == p1[j - 1]) {
kind[5][1] = p1[i];
if (len1 == 5)
only[5][1] = 1;
break;
}
}
vis[i - 2] = 0;
vis[i - 1] = 0;
vis[i] = 0;
if (kind[5][1] != -1)
break;
}
}
if (only[5][1]) {
printf("Yes\n");
continue;
}
memset(vis, 0, sizeof(vis));
for (i = 2; i < len2; i++) {
if (p2[i - 1] == p2[i] && p2[i - 2] == p2[i - 1]) {
vis[i - 2] = 1;
vis[i - 1] = 1;
vis[i] = 1;
for (j = 1; j < len2; j++) {
if ((!vis[j - 1]) && (!vis[j]) && p2[j] == p2[j - 1]) {
kind[5][2] = p2[i];
break;
}
}
vis[i - 2] = 0;
vis[i - 1] = 0;
vis[i] = 0;
if (kind[5][2] != -1)
break;
}
}
if (kind[5][1] != -1 && kind[5][1] >= kind[5][2]) {
if ((kind[7][2] == -1) && (kind[8][2] == -1)) {
printf("Yes\n");
continue;
}
}
//6
for (i = 3; i < len1; i++) {
if (p1[i - 3] == p1[i - 2] && p1[i - 2] == p1[i - 1]
&& p1[i - 1] == p1[i]) {
if (len1 >= 6) {
kind[6][1] = p1[i];
if (len1 == 6)
only[6][1] = 1;
break;
}
}
}
if (only[6][1]) {
printf("Yes\n");
continue;
}
for (i = 3; i < len2; i++) {
if (p2[i - 3] == p2[i - 2] && p2[i - 2] == p2[i - 1]
&& p2[i - 1] == p2[i]) {
if (len2 >= 6) {
kind[6][2] = p2[i];
break;
}
}
}
if (kind[6][1] != -1 && kind[6][1] >= kind[6][2]) {
if ((kind[7][2] == -1) && (kind[8][2] == -1)) {
printf("Yes\n");
continue;
}
}
printf("No\n");
}
return 0;
}
2014多校联合六（HDU 4923 HDU 4925 HDU 4927 HDU 4930）,布布扣,bubuko.com
2014多校联合六（HDU 4923 HDU 4925 HDU 4927 HDU 4930）
原文地址：http://blog.csdn.net/houserabbit/article/details/38423849
