CodeForces - 552A Vanya and Table （水）

CodeForces - 552A Vanya and Table Time Limit: 2000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status 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 rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result. 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 are the number of the column and row of the upper right cell of a rectangle. Output In a single line print the sum of all values in the cells of the table. Sample Input Input 2 1 1 2 3 2 2 3 3 Output 10 Input 2 1 1 3 3 1 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; }