JZOJ 5812. 【NOIP提高A组模拟2018.8.14】区间

5812. 【NOIP提高A组模拟2018.8.14】区间
(File IO): input:range.in output:range.out

Time Limits: 1000 ms Memory Limits: 262144 KB Detailed Limits Special Judge Downloads

Description

每一个机房中总有一个红太阳。有一天，AmberFrame 来到机房，发现桌上有不知道哪个蒟蒻放上的问题：有一个 n 个数的序列，一开始所有的数都是 0，每次可以将一个区间 [l, r](l ≤ r) 内的数 +1，求到达最终状态的最少操作次数。 AmberFrame 非常强，自然不会把时间花在这种水题上。因此他就把任务交给了你，如果不会做的话，他可能就会觉得你就是那个放问题的蒟蒻了而把你批判一番了。

Input

第一行包含一个正整数 n，表示序列的长度。
第二行包含 n 个非负整数 a1, a2, ..., an，表示最终的状态。

Output

输出的第一行是一个正整数 m，表示最少的操作次数。接下来 m 行每行两个正整数 li , ri，表示一次操作。你需要保证 1 ≤ li ≤ ri ≤ n。保证最少次数 m ≤ 10^5，输出可以以任意顺序输出。

Sample Input

6

2 3 3 3 3 3

Sample Output

Data Constraint

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" alt=" " />

Hint

下发样例中第 i 个样例与第 i 组数据范围相符。
对于样例 1，第一个数被加了两次，其他每个数都被加了三次，显然满足条件。

做法：一开始还想用线段树来着，后来发现会超时，题解上说打差分从头到尾扫一遍就好啦

代码如下：

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #define N 100007

 using namespace std;

 int n, a[N], list[N], tot, ans;

 int abs (int x) { return x >  ? x : -x; }

 void Work(int step, int Times)

 {

     for (; Times--; list[++tot] = step);

 }

 void Print(int step, int Times)

 {

     for(; Times--; printf("%d %d\n", list[tot--], step));

 }

 int main()

 {

     freopen("range.in", "r", stdin);

     freopen("range.out", "w", stdout);

     scanf("%d", &n);

     for (int i = ; i <= n; scanf("%d", &a[i++]));

     for (int i = ; i <= n; i++)

         if (a[i] > a[i - ])    ans += a[i] - a[i - ];

     printf("%d\n", ans);

     for (int i = ; i <= n + ; i++)

     {

         if (a[i] > a[i - ])     Work(i, abs(a[i] - a[i - ]));

         if (a[i] < a[i - ])    Print(i - , abs(a[i] - a[i - ]));

     }

 }