CodeForces336 A & B

Posted 2020-06-09

第一题就是排序然后计算一下时间。没什么

package codeforces336;

import java.io.InputStreamReader;
import java.util.Scanner;

public class MainA {
    public static void sortArr(int[][] arr, int n) {
        int[] tmp = new int[2];
        for(int i = 0; i < n; ++i) {
            for(int j = i+1; j < n; ++j) {
                if(arr[i][0] < arr[j][0]) {
                    tmp[0] = arr[i][0];
                    tmp[1] = arr[i][1];
                    arr[i][0] = arr[j][0];
                    arr[i][1] = arr[j][1];
                    arr[j][0] = tmp[0];
                    arr[j][1] = tmp[1];
                }
            }
        }
    }

    public static void main(String[] args) {
        int n, s;
        int[][] arr = new int[105][2];
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        n = sc.nextInt();
        s = sc.nextInt();
        for(int i = 0; i < n; ++i) {
            arr[i][0] = sc.nextInt();
            arr[i][1] = sc.nextInt();
        }

        sortArr(arr, n);
        int t = 0;
        int[] tmp = new int[2];
        for(int i = 0; i < n; ++i) {
            tmp[0] = s - arr[i][0];
            tmp[1] = arr[i][1];
            if(tmp[1] < (t + tmp[0])){
                t += tmp[0];
            } else {
                t = tmp[1];
            }
            s = arr[i][0];
        }
        if(s != 0) {
            t += s;
        }
        System.out.println(t);
    }
}

 第二题，暴力肯定TLE，用前缀和算可以。看a的每一位，是0，统计在 lenb - lena + i  ~ i - 1 范围内 1的 个数；是 1，统计在 lenb - lena + i  ~ i - 1 范围内 0 的 个数

package codeforces336;

import java.io.InputStreamReader;
import java.util.Scanner;

/**
 * Created by lenovo on 2016-01-28.
 */
public class MainB {
    public static void main(String[] args) {
        String a;
        String b;
        Scanner sc = new Scanner(new InputStreamReader(System.in));
        a = sc.nextLine();
        b = sc.nextLine();
        //System.out.println(a + " " + b);
        long[][] pre = new long[200005][2];
        int lena = a.length();
        int lenb = b.length();

        for(int i = 1; i <= lenb; ++i) {
            if(b.charAt(i-1) == ‘1‘){
                pre[i][1] = pre[i-1][1] + 1;
                pre[i][0] = pre[i-1][0];
            }else {
                pre[i][0] = pre[i-1][0] + 1;
                pre[i][1] = pre[i-1][1];
            }
        }
        long  ans = 0;
        for(int i = 1; i <= lena; ++i) {
            if(a.charAt(i-1) == ‘0‘) {
                ans += pre[lenb - lena + i][1] - pre[i-1][1];
            } else {
                ans += pre[lenb - lena + i][0] - pre[i-1][0];
            }
        }
        System.out.println(ans);
    }
}