codeforces Educational Codeforces Round 16 D. Two Arithmetic Progressions 扩展欧几里得

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先用exgcd等式求k的通解然后根绝函数的单调性确定可选整数的区间注意点很多有点恶心
然后求个数就行了
坑点是整数运算两个数相除 a/b 的结果为正返回小于等于结果的第一个数，结果为负返回大于等于结果的第一个数

//1 9 3 11 49 109#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#define lson i<<1#define rson (i<<1)+1#define LL long long#define pi acos(-1.0)#define mid (l+r)/2#define maxn 10000005using namespace std;LL x,y;LL e_gcd(LL a, LL b){    if(b==0)    {       x = 1;       y = 0;       return a;    }    LL ans = e_gcd(b,a%b);    LL t = x;    x = y;    y = t-a/b*y;    return ans;}  //求通解LL a1,b1,a2,b2,l,r;int main(){    scanf("%I64d %I64d %I64d %I64d %I64d %I64d", &a1, &b1, &a2, &b2, &l, &r);    //扩展欧几里得求通解    LL a = a1, b = -1 * a2;    LL GCD = e_gcd(a, b);    if((b2 - b1)%GCD != 0) printf("0\n");    else{        //求k左右边界        x *= (b2 - b1)/GCD;        LL add = b / GCD ;        add = abs(add);        if((l - b1) % a1==0) l = (l - b1) / a1;        else if(l - b1 >= 0)l = (l - b1) / a1 + 1;        else l = (l - b1) / a1 ;        if((r - b1) % a1==0) r = (r - b1) / a1;        else if(r - b1 >= 0)r = (r - b1) / a1;        else r = (r - b1) / a1 - 1;        //根据l>=0的条件刷新边界 且k>=0刷新边界        if(r < 0) { printf("0\n"); return 0; }        if(r * a1 + b1 - b2 < 0) { printf("0\n"); return 0;}        l = max(0ll, l);        if(l * a1 + b1 - b2 < 0)        {            if((b2 - b1) % a1 != 0) l = (b2 - b1) / a1 + 1;            else  l = (b2 - b1) / a1;        }        //求l,r内的解        l-=x;        r-=x;        l += (add - (l % add + add) % add) % add;        r -= (r % add + add) % add;        printf("%I64d\n", max(0ll, (r - l)/add + 1));    }    return 0;}

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