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);
}
}