2014多校聯合六（HDU 4923 HDU 4925 HDU 4927 HDU 4930）

標籤：hdu   數學   雜題   

HDU 4923 Room and Moor

題意：給出A序列  求滿足題目所寫的B序列  使得方差最小

思路：可以想到最後的結果中  B序列的值一定是一段一段的  那麼我們可以類似貪心去搞  對於一段序列我們可以求出什麼樣的b值使得方差最小  即序列中1的個數除以序列長度  又因為B是單調的  可以用一個單調棧去類比  複雜度遠遠小於n^2  不要被嚇怕…

代碼：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 100010int 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

題意：n*m的格子  要麼種蘋果  要麼施化肥  施肥後的格子的相鄰格子如果種了蘋果  則蘋果數翻倍  問最多產生幾個蘋果

思路：明顯就是按矩形黑白染色的方法來種樹和施肥  那麼枚舉兩種情況  即是黑格子還是白格子種蘋果  最後取max

代碼：

#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

題意：一個序列A  每次根據這個方程ai=ai+1-ai使整個序列數字個數-1  問n-1次操作後剩下哪個數

思路：水題  只要把式子化簡即可  注意高精度

代碼：

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

題意：鬥地主- -b  如果一下把手牌出光就算贏  或者  你出一次對手管不上就算贏  問能不能贏

思路：類比題…  細心就好   邏輯如下

先看是否一次能出完  如果不能  判斷大小王  在自己手裡必勝在別人手裡必輸

如果還沒有結果   判炸彈  如果還沒結果  就分所有能出牌的可能判斷

代碼：

#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;elsereturn 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);//7if (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;}//8for (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;}//1kind[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;}}//2for (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;}}//3for (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;}}//4for (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;}}//5memset(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;}}//6for (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;}