A题：
一道dp的题目
dp[i][j] = k 代表 i行放j个棋子有k中可能
dp[i][j] = dp[i-1][0] + dp[i-1][1] + dp[i-1][2] +...dp[i-1][j]
初始 dp[1][0] = 1, dp[1][1] = 1
注意点：
- BigInteger 否则爆栈
- 最后return 从1开始加
import java.util.*;
import java.math.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
BigInteger[][] dp = new BigInteger[110][110];
for (int i = 0;i <= 109; i++)
for(int j = 0;j <= 109; j++)
dp[i][j] = new BigInteger("0");
dp[1][0] = new BigInteger("1");
dp[1][1] = new BigInteger("1");
int n = sc.nextInt();
for (int i = 2; i <= n ;i++){
for (int j = 0; j <= i; j++){
for (int k = 0; k <= j; k++){
dp[i][j] = dp[i][j].add(dp[i-1][k]);
}
}
}
BigInteger ans = new BigInteger("0");
for (int i = 1; i <= n; i++)
ans = ans.add(dp[n][i]);
System.out.println(ans);
}
}
B题：用公式
不知道这个A的伴随矩阵公式，就很难受了- - 。。
import java.util.*;
public class Main2{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for (int i = 0; i < T; i++) {
int[][] A = new int[3][3];
for (int m = 0; m < 3; m ++) {
for (int n = 0; n < 3; n++) {
A[m][n] = sc.nextInt();
}
}
long ao = (A[0][0]* A[1][1]* A[2][2]) + (A[1][0]* A[2][1]* A[0][2]) + (A[2][0]* A[0][1]* A[1][2])
- (A[0][2]* A[1][1]* A[2][0]) - (A[0][0]* A[1][2]* A[2][1]) - (A[0][1]* A[1][0]* A[2][2]);
System.out.println(ao* ao);
}
}
}
C题：
一开始理解错了，他的意思是a^n，我还以为要a *=a;
import java.math.BigInteger;
import java.util.*;
public class Main3{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for (int i = 0; i < T; i++) {
BigInteger a = new BigInteger("0"), n = new BigInteger("0"), b = new BigInteger("0");
a = sc.nextBigInteger();
n = sc.nextBigInteger();
b = sc.nextBigInteger();
System.out.println(a.modPow(n,b));
}
}
}
E题：
感觉巨他妈坑了，本来
××if (y1_xing == C)
××B --;
我都服了，被这个卡了小半天。。
import java.math.BigInteger;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.*;
public class Main{
static int y1_xing;
static boolean r;
static int ny, nm, nd;
static int[] monthofday = new int[]{-1,31,28,31,30,31,30,31,31,30,31,30,31};
static int res; //这个月的几号
public static boolean is_run(int year) {
if (year % 400 == 0 ||(year % 4 == 0 && year % 100 != 0))
return true;
return false;
}
public static void gety1_xing(int ny,int y,int A, int B, int C) {
y1_xing = 1;
for (ny = 1850; ny < y; ny++) {
for (nm = 1; nm <= 12; nm ++) {
int month_ofday = monthofday[nm];
if(is_run(ny) && nm == 2) month_ofday += 1;
for(nd = 1; nd <= month_ofday; nd++) {
y1_xing += 1;
if (y1_xing == 8) y1_xing = 1;
}
}
}
for(nm = 1;nm < A;nm++) {
int month_ofday = monthofday[nm];
if(is_run(y) && nm == 2) month_ofday += 1;
for(nd = 1; nd <= month_ofday; nd++) {
y1_xing += 1;
if (y1_xing == 8) y1_xing = 1;
}
}
boolean has = false;
int month_ofday = monthofday[nm];
if(is_run(y) && B == 2) month_ofday += 1;
for (res = 1;res <= month_ofday; res++) {
y1_xing += 1;
if (y1_xing == 8) {
y1_xing = 1;
}
if (y1_xing == C)
B --;
if (B == 0) {
has = true;
break;
}
}
if(!has) res = -1;
}
public static void main(String[] args) throws ParseException{
Scanner sc = new Scanner(System.in);
while (sc.hasNextInt()) {
int A = sc.nextInt(), B = sc.nextInt(), C = sc.nextInt(), y = sc.nextInt();
gety1_xing(ny,y, A, B, C);
if (res != -1) {
Calendar c = Calendar.getInstance();
c.set(Calendar.YEAR, y);
c.set(Calendar.MONTH, A - 1);
c.set(Calendar.DAY_OF_MONTH, res);
Date d=new Date();
d=c.getTime();
SimpleDateFormat sdf=new SimpleDateFormat("yyyy/MM/dd");
String str=sdf.format(d);
System.out.println(str);
}
else
System.out.println("none");
}
}
}
H题：
先打表，在找规律！
import java.math.BigInteger;
import java.util.*;
public class MainH{
//打表 -> 找规律！
//100 1-1 1-2 1-3 1-4....1-100
//99 1-1 1-2 1-3...1-99
//...
//2 1-1 1-2
//1 1-1
public static void main(String[] args){
/*
for(int i = 1; i <= 10 ; i++) {
int[] a = new int[i + 1]; //计数从1开始
//经过i次
for(int j = 2; j <= i; j++) {
for(int k = 1; k*j < a.length; k++) {
a[k*j] = 1 - a[k*j];
}
}
for (int j = 1; j <= i; j++) {
System.out.print(a[j]+" ");
}
System.out.println();
}*/
//找到规律：规律是：
//经过i次，为1的位置为：1 4 9 16 25 36 49 64 81 100 121 144
// A=1,B=3
//4 - 1 - 1 = 2, 9 - 4 - 1 =4,
//16 - 9 - 1 = 6, 25 - 16 - 1 = 8
//依次增长2
//1: 0 0 0 0 0
//2: 0 1 0 1 0
//3: 0 1 1 1 0
//4: 0 1 1 0 0
//5: 0 1 1 0 1
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for (int i = 0; i < T; i++) {
long N = sc.nextLong(), A = sc.nextLong(),B = sc.nextLong();
int start = 1;
int gap = 2;
while (start < A) {
start += gap + 1;
gap += 2;
}
//start >= A
int res = 0;
while (start <= B) {
res ++;
start += gap + 1;
gap += 2;
}
System.out.println(B - A+1-res);
}
}
}
感觉没有练习过ACM 的确是做得磕磕拌拌，全是坑！！！ 明天接着更剩下的题吧- - 。。