789. 数的范围 - AcWing题库 

import java.util.*;public class Main{static int N = 100010;static int[] a = new int[N];public static void main(String[] args){Scanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();for(int i = 0; i < n; i ++){a[i] = sc.nextInt();}while(m -- > 0){int x = sc.nextInt();int l = 0, r = n - 1;while(l < r){int mid = l + r >> 1;if(a[mid] >= x) r = mid;//找到区间的左端点（区分大于等于x的和小于x的）else l = mid + 1;}if(a[l] != x) System.out.println("-1 -1");else{System.out.print(l + " ");l = 0;r = n - 1;while(l < r){int mid = l + r + 1 >> 1;if(a[mid] <= x) l = mid;//找到区间的右端点（区分小于等于x的和大于x的）else r = mid - 1;}System.out.println(l);}}}
}

790. 数的三次方根 - AcWing题库

import java.util.*;public class Main{public static void main(String[] args){Scanner sc = new Scanner(System.in);double x = sc.nextDouble();double l = -100, r = 100;while(r - l > 1e-8){double mid = (l + r) / 2;if(mid * mid * mid >= x) r = mid;else l = mid;}System.out.printf("%.6f", l);}
}

 795. 前缀和 - AcWing题库

import java.util.*;public class Main{public static void main(String[] args){Scanner sc = new Scanner(System.in);int N = 100010;int[] a = new int[N];int[] b = new int[N];int n = sc.nextInt();int m = sc.nextInt();for(int i = 1; i <= n; i ++){a[i] = sc.nextInt();//数组的每个数}for(int i = 1; i <= n; i ++){b[i] = b[i - 1] + a[i];//前缀和}while(m -- > 0){int l = sc.nextInt();int r = sc.nextInt();System.out.println(b[r] - b[l - 1]);}}
}

796. 子矩阵的和 - AcWing题库

import java.util.*;public class Main{static int N = 1010;static int[][] a = new int[N][N];//每个位置的值static int[][] s = new int[N][N];//二维前缀和public static void main(String[] args){Scanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int q = sc.nextInt();for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){a[i][j] = sc.nextInt();}}for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];}}while(q -- > 0){int x1 = sc.nextInt();int y1 = sc.nextInt();int x2 = sc.nextInt();int y2 = sc.nextInt();System.out.println(s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);}}
}

1227. 分巧克力 - AcWing题库

import java.util.*;public class Main{static int N = 100010;static int n, m;static int[] h = new int[N];static int[] w = new int[N];public static boolean check(int x){int res = 0;for(int i = 1; i <= n; i ++){res += (h[i] / x) * (w[i] / x);}if(res >= m) return true;return false;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();int max = 0;for(int i = 1; i <= n; i ++){h[i] = sc.nextInt();w[i] = sc.nextInt();max = Math.max(h[i], max);max = Math.max(w[i], max);}int l = 0, r = max;while(l < r){int mid = l + r + 1>> 1;if(check(mid)) l = mid;else  r = mid - 1;}System.out.print(l);}
}

 

1230. K倍区间 - AcWing题库

import java.util.*;public class Main{static int N = 100010;static long[] a = new long[N];static long[] s = new long[N];//前缀和static long[] cnt = new long[N];//模k余数为xstatic int n, k;public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();k = sc.nextInt();for(int i = 1; i <= n; i ++){a[i] = sc.nextInt();}for(int i = 1; i <= n; i ++){s[i] = s[i - 1] + a[i];}long res = 0;cnt[0] = 1;//初始化余数为0也是一种情况for(int i = 1; i <= n; i ++){res += cnt[(int)(s[i] % k)];//在还没有改变模k的余数为x的前缀和的个数之前，加到答案中cnt[(int)(s[i] % k)] ++;//个数加1}System.out.print(res);}
}

 

730. 机器人跳跃问题 - AcWing题库

import java.util.*;public class Main{static int N = 100010;static int[] h = new int[N];static int n, max;public static boolean check(int x){for(int i = 1; i <= n; i ++){x = 2 * x - h[i];if(x >= max) return true;//只要存在能量大于等于最大值，就已经可以保证了if(x < 0) return false;}return true;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();max = 0;for(int i = 1; i <= n; i ++){h[i] = sc.nextInt();max = Math.max(max, h[i]);}int l = 0, r = N;while(l < r){int mid = l + r >> 1;if(check(mid)) r = mid;else l = mid + 1;}System.out.print(l);}
}

 

1221. 四平方和 - AcWing题库

import java.util.*;class PII implements Comparable<PII>{int a, b, c;public PII(int a, int b, int c){this.a = a;this.b = b;this.c =c;}public int compareTo(PII o){if(a != o.a) return Integer.compare(a, o.a);if(b != o.b) return Integer.compare(b, o.b);return Integer.compare(c, o.c);}
}public class Main{static int N = 2500010;static PII[] sum = new PII[N];static int n, m;public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();for(int c = 0; c * c <= n; c ++){for(int d = c; d * d + c * c <= n; d ++){sum[m ++] = new PII(d * d + c * c, c, d);}}Arrays.sort(sum, 0, m);//按照定义的次序排序for(int a = 0; a * a <= n; a ++){for(int b = a; b * b + a * a <= n; b ++){int t = n - (a * a + b * b);//通过二分判断这个t是否在sum数组中出现过int l = 0, r = m - 1;while(l < r){int mid = l + r >> 1;if(sum[mid].a >= t) r = mid;else l = mid + 1;}if(sum[l].a == t){System.out.print(a + " " + b + " " + sum[l].b + " " + sum[l].c);return;} }}}
}