CodeForces - 552A Vanya and Table （水）

64bit IO Format: %I64d & %I64u

CodeForces - 552A

Vanya and Table

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Description

Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100from left to right.

In this table, Vanya chose n

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of rectangles.

Each of the following n lines contains four integers x₁, y₁, x₂, y₂ (1 ≤ x₁ ≤ x₂ ≤ 100, 1 ≤ y₁ ≤ y₂ ≤ 100), where x₁ and y₁ are the number of the column and row of the lower left cell and x₂ and y₂

Output

In a single line print the sum of all values in the cells of the table.

Sample Input

Input

21 1 2 32 2 3 3

Output

10

Input

21 1 3 31 1 3 3

Output

18

Hint

Note to the first sample test:

Values of the table in the first three rows and columns will be as follows:

121

121

110

So, the sum of values will be equal to 10.

Note to the second sample test:

Values of the table in the first three rows and columns will be as follows:

222

222

222

So, the sum of values will be equal to 18.

Source

Codeforces Round #308 (Div. 2)

//题意：输入一个n，接下来输入n个（x1,y1）(x2,y2).

表示给你n个矩形的左下（横纵坐标都加了1，计算时减去即可）和右上的坐标，让你计算这n个矩形的面积。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
	int n,i,j;
	int sum;
	int x1,x2,y1,y2;
	while(scanf("%d",&n)!=EOF)
	{
		sum=0;
		while(n--)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			sum+=(x2-x1+1)*(y2-y1+1);
		}
		printf("%d\n",sum);
	}
	return 0;
}