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);    }}