BZOJ5027: mathematical questions, bzoj5027 mathematical questions

BZOJ5027: mathematical questions, bzoj5027 mathematical questions

Time Limit: 10 Sec Memory Limit: 256 MB
Submit: 140 Solved: 48
[Submit] [Status] [Discuss] Description: returns a, B, c, x1, x2, y1, y2. Obtain the conditions that meet the requirements of ax + by + c = 0, and x in [x1, x2], y in [y1, y2] How many pairs of integer solutions are there?

 

The first line of Input contains seven integers, a, B, c, x1, x2, y1, y2, which are separated by spaces. The absolute values of a, B, c, x1, x2, y1, y2 cannot exceed 10 ^ 8.

 

How many pairs of Output integer solutions?

 

Sample Input1 1-3 0 4 0 4 Sample Output4HINT Source It can be seen at a glance that it is Europe Expansion Process the upper and lower boundary using the general formula of the expanded Europe Region Pay attention to the judgment in special cases Note:

Must first multiply and then divide

No error occurred when calling the MPs for one night and taking n groups of data ..

 

# Include <iostream> # include <cstdio> # include <cmath> # define LL long using namespace std; const ll maxn = 1e6 + 10; LL a, B, c, x1, x2, yy1, y2, x, y; LL exgcd (LL a, LL B, LL & x, LL & y) {if (B = 0) {x = 1, y = 0; return a;} LL r = exgcd (B, a % B, x, y), tmp; tmp = x, x = y, y = tmp-a/B * y; return r;} LL min (LL a, LL B) {return a <B? A: B;} LL max (LL a, LL B) {return a> B? A: B;} int main () {cin> a> B> c> x1> x2> yy1> y2; c =-c; if (a = 0 & B = 0) {if (c = 0) {printf ("% lld", (LL) (x2-x1 + 1) * (y2-yy1 + 1); return 0;} else {printf ("0"); return 0 ;}} if (a = 0) {if (c % B) {printf ("0"); return 0;} if (c/B> = yy1 & c/B <= y2) {printf ("% lld ", x2-x1 + 1); return 0;} else {printf ("0"); return 0 ;}}if (B = 0) {if (c %) {printf ("0"); return 0;} if (c/a> = x1 & c/a <= x2) {printf ("% lld ", y2-yy1 + 1); return 0;} else {printf ("0"); return 0 ;}} LL r = exgcd (a, B, x, y ); B = B/r; a =-a/r; // use the formula to construct the incremental if (c % r) {printf ("0"); return 0 ;} x = x * c/r; y = y * c/r; LL xlower, xupper, ylower, yupper; if (B> 0) xlower = ceil (double) (x1-x)/B), xupper = floor (double) (x2-x)/B); if (B <0) xlower = ceil (double) (x2-x) /B), xupper = floor (double) (x1-x)/B); if (a> 0) ylower = ceil (double) (yy1-y)/), yupper = floor (double) (y2-y)/a); if (a <0) ylower = ceil (double) (y2-y)/), yupper = floor (double) (yy1-y)/a); LL ans = max (0, min (xupper, yupper)-max (xlower, ylower) + 1 ); printf ("% lld", ans); return 0;} // 1 5-3-123 40-567 41