题目相关

题目链接

AtCoder Beginner Contest 181 B 题，https://atcoder.jp/contests/abc181/tasks/abc181_b。

Problem Statement

We have a blackboard with nothing written on it. Takahashi will do N operations to write integers on it.

In the i-th operation, he will write each integer from Ai through Bi once, for a total of Bi−Ai+1 integers.

Find the sum of the integers written on the blackboard after the N operations.

Input

Input is given from Standard Input in the following format:

N
A1 B1
.
.
.
AN BN

Output

Print the sum of the integers written on the blackboard after the N operations.

Samples1

Sample Input 1

2
1 3
3 5

Sample Output 1

Explaination

In the 1-st operation, he will write 1, 2, and 3 on the blackboard.

In the 2-nd operation, he will write 3, 4, and 5 on the blackboard.

The sum of the integers written is 1+2+3+3+4+5=18.

Samples2

Sample Input 2

Sample Output 2

998244353

Samples3

Sample Input 3

1
1 1000000

Sample Output 3

500000500000

Constraints

All values in input are integers.
1≤N≤10^5
1≤Ai≤Bi≤10^6

题解报告

题目翻译

高桥在白板写出 N 行，每行包括两个整数。第 i 行操作，将写出数据 Ai 和 Bi，这样一共有 Bi-Ai+1 个数字。请计算出这 N 次操作包含的所有数据总和。

题目分析

回到了 AtCoder Beginner Contest 后，感觉明显好多了。看来自己的水平也就是这样。

我们可以知道每次高桥写出一个公差为 1，数量为 Bi-Ai+1 个数字。我们可以根据等差数量求和公式： $S=n*A_{i}+\frac{n*(n-1)}{2}*d$ 来计算。

数据规模估计

从数据限制可知，最终的总和最大值为： $\frac{10^6*(10^6+1)}{2}*10^5\approx 10^6*10^6*10^5=10^{17}$ 。说明需要使用 long long 来表示。

AC 参考代码

//https://atcoder.jp/contests/abc181/tasks/abc181_b
//B - Trapezoid Sum
//d=1的等差数列求和
#include <iostream>
using namespace std;
int main() {int n;cin>>n;long long ans=0;while (n--) {long long a,b;cin>>a>>b;long long tot=b-a+1;ans+=(a*tot+tot*(tot-1)/2);}cout<<ans<<"\n";return 0;
}

时间复杂度

O(N)。

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