第一題:編碼數目
題目:輸入N,M,求N+N^2+N^3+...+N^M的結果(取餘1000000007),1<N<=65536,1<M<=100000
輸入格式:每行輸入N M,直到N M均等於0時跳出
輸出格式:每行輸出對應的結果
解題思路:快速冪(實質對冪指數進行二進制轉換,將累乘拆分),不瞭解的可以查找一下
import java.util.*;
public class Main {
static Scanner in = new Scanner(System.in);
static int mod = 1000000007;
static int q_pow(int a, int b) {
a %= mod;
int ans = 1;
while (b != 0) {
if ((b & 1) == 1)
ans = (ans * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return ans;
}
public static void main(String[] args) {
int n = in.nextInt();
int m = in.nextInt();
long sum = 0;
for (int i = 1; i <= m; i++) {
sum+=q_pow(n, i);//累加即可
}
System.out.println(sum);
}
}
其實還有一種思路是利用Java提供的大數類BigInterger,這個類支持大數計算
第二題:二進制
題目:允許對二進制數進行兩種操作:00->10,10->01,求可能的最大數(兩種操作可以進行任意次)
輸入格式:先一行輸出樣例數,然後每兩行輸入二進制長度與二進制數本體,1<=長度<=10000
例如:
2
4
0001
10
1010111000
輸出格式:每行輸出每個樣例的答案
例如:
1101
1111101111
思路:操作:①00->10②10->01,
爲了使得到的數最大,儘可能使高位爲1,那麼②其實是爲①服務的,利用②將0往高位移動,將1往低位移動,然後將高位的0利用①轉換,通過模擬可以得到結論:第一個0後面的所有1(設個數爲a)都會別挪到最後(通過②),此時經過①會有新的0101情況出現,繼續重複上述步驟,最後得到:(總長度-a-1)個1+0+a個1 的字符串
總結思路:(1)通過②將第一個0後面的所有1移到最後
(2)通過①處理此時的字符串
(3)重複(1)(2)直到不能進行①②操作
import java.util.*;
public class Main {
static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
int t = in.nextInt();
while(t-->0) {
int n = in.nextInt();
String s = in.next();
int pos = 0;
int cnt = 0;
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i)=='0') {pos = i;break;};
}
for (int i = pos; i < s.length(); i++) {
if(s.charAt(i)=='1') cnt++;
}
String ans = "";
for (int i = 0; i < n-cnt-1; i++) {
ans+="1";
}
ans+="0";
for (int i = 0; i < cnt; i++) {
ans+="1";
}
if(pos==0)//注意字符串全爲1的情況
System.out.println(s);
else
System.out.println(ans);
}
}
}
第三題:解數獨
題目:即leetcode 37題
輸入格式:
{5,3,0,0,7,0,0,0,0}
{6,0,0,1,9,5,0,0,0}
{0,9,8,0,0,0,0,6,0}
{8,0,0,0,6,0,0,0,3}
{4,0,0,8,0,3,0,0,1}
{7,0,0,0,2,0,0,0,6}
{0,6,0,0,0,0,2,8,0}
{0,0,0,4,1,9,0,0,5}
{0,0,0,0,8,0,0,7,9}
輸出格式:
{5,3,4,6,7,8,9,1,2}
{6,7,2,1,9,5,3,4,8}
{1,9,8,3,4,2,5,6,7}
{8,5,9,7,6,1,4,2,3}
{4,2,6,8,5,3,7,9,1}
{7,1,3,9,2,4,8,5,6}
{9,6,1,5,3,7,2,8,4}
{2,8,7,4,1,9,6,3,5}
{3,4,5,2,8,6,1,7,9}
思路:標誌數組+回溯
import java.util.Scanner;
public class Main {
static Scanner in = new Scanner(System.in);
static boolean[][] row = new boolean[9][10];
static boolean[][] col = new boolean[9][10];
static boolean[][] box = new boolean[9][10];
static int[][] num = new int[9][9];
static boolean solve(int r, int c) {
if (c == 9) {
c = 0;
r++;
if (r == 9)
return true;
}
if (num[r][c] == 0) {
int id = 0;
for (int i = 1; i <= 9; i++) {
id = (r / 3) * 3 + c / 3;
//不能在同一行同一列或者同一個方框
if (row[r][i] == false && col[c][i] == false && box[id][i] == false) {
row[r][i] = col[c][i] = box[id][i] = true;
num[r][c] = i;
if (solve(r, c + 1))
return true;
//位置不合適,回溯
num[r][c] = 0;
row[r][i] = col[c][i] = box[id][i] = false;
}
}
}else
return solve(r, c + 1);
return false;
}
public static void main(String[] args) {
String str;
int bid = 0;
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 10; j++) {
row[i][j] = col[i][j] = box[i][j] = false;
}
}
for (int i = 0; i < 9; i++) {
str = in.next();
for (int j = 0; j < 9; j++) {
num[i][j] = str.charAt(j * 2 + 1) - '0';
if (num[i][j] >= 1 && num[i][j] <= 9) {
bid = (i / 3) * 3 + j / 3;//每個3X3方塊的索引
row[i][num[i][j]] = true;
col[j][num[i][j]] = true;
box[bid][num[i][j]] = true;
}
}
}
solve(0, 0);
for (int i = 0; i < 9; i++) {
System.out.print("{");
for (int j = 0; j < 9; j++) {
if (j != 8)
System.out.print(num[i][j] + ",");
else
System.out.println(num[i][j] + "}");
}
}
}
}