import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
int n = in.nextInt(),q = in.nextInt();
int[] arr = new int[n+1];
for(int i = 1; i < n+1; i++){
arr[i] = in.nextInt();
}
long[] dp = new long[n+1];
for(int i = 1; i < n+1; i++){
dp[i] = dp[i-1] + arr[i];
}
while(q > 0){
int l = in.nextInt(), r = in.nextInt();
System.out.println(dp[r] - dp[l-1]);
q--;
}
// while (in.hasNextInt()) { // 注意 while 处理多个 case
// int a = in.nextInt();
// int b = in.nextInt();
// System.out.println(a + b);
// }
}
}
DP35 【模板】二维前缀和
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt(), m = in.nextInt(), q = in.nextInt();
int[][] arr = new int[n+1][m+1];
for(int i = 1; i < n+1; i++){
for(int j = 1; j < m+1; j++){
arr[i][j] = in.nextInt();
}
}
long[][] dp= new long[n+1][m+1];
for(int i = 1; i < n+1; i++){
for(int j = 1; j < m+1; j++){
dp[i][j] = dp[i][j-1] + dp[i-1][j] + arr[i][j] - dp[i-1][j-1];
}
}
while(q > 0){
int x1 = in.nextInt(),y1 = in.nextInt(),
x2 = in.nextInt(),y2 = in.nextInt();
System.out.println(dp[x2][y2]-dp[x1-1][y2]
- dp[x2][y1-1] + dp[x1-1][y1-1]);
q--;
}
// 注意 hasNext 和 hasNextLine 的区别
}
}
724. 寻找数组的中心下标
class Solution {
public int pivotIndex(int[] nums) {
int[] dp = new int[nums.length];
dp[0] = nums[0];
for(int i = 1; i < nums.length; i++){
dp[i] = dp[i-1] + nums[i];
}
if(0 == dp[nums.length-1] -dp[0] ){
return 0;
}
for(int i = 1; i < nums.length; i++){
if(dp[i-1] == dp[nums.length-1] - dp[i]){
return i;
}
}
return -1;
}
}
238. 除自身以外数组的乘积
class Solution {
public int[] productExceptSelf(int[] nums) {
int l = nums.length;
int[] f = new int[l];
int[] g = new int[l];
f[0] = 1;
g[l-1] = 1;
for(int i = 1; i < l; i++){
f[i] = f[i-1]*nums[i-1];
}
for(int i = l-2; i >= 0; i--){
g[i] = g[i+1]*nums[i+1];
}
int[] answer = new int[l];
for(int i = 0; i < l; i++){
answer[i] = f[i]*g[i];
}
return answer;
}
}