标签：typedef scan freopen else 恶心 exgcd include mes mat
各种傻逼特判，判错一个就40
思路有两种
首先求出来左右边界然后拿右边界减左边界，非常简单，按照解方程的方法求即可
代码（我没打出来一直40分，这里是nc的代码）
#include<iostream> #include<cstdio> #include<cmath> using namespace std; typedef long long ll; ll T,a,b,c,x,y; ll exgcd(ll a,ll b,ll &x,ll &y) { if(b==0) { x=1;y=0; return a; } ll d=exgcd(b,a%b,x,y); ll tmp=x; x=y; y=tmp-a/b*y; return d; } int main() { scanf("%lld",&T); while(T--) { scanf("%lld%lld%lld",&a,&b,&c); ll d=exgcd(a,b,x,y); ll l=floor(1.0*c/b*(-x)),r=ceil(1.0*c/a*y); ll k=r-l-1; if(a==0&&b==0) { if(c==0) { puts("ZenMeZheMeDuo"); continue; } else { puts("0"); continue; } } if((c%d)) { puts("0"); continue; } if(1LL*a*b<0) { puts("ZenMeZheMeDuo"); continue; } if(a==0) { if(1LL*b*c>0) puts("ZenMeZheMeDuo"); else puts("0"); continue; } if(b==0) { if(1LL*a*c>0) puts("ZenMeZheMeDuo"); else puts("0"); continue; } if(k<0) { puts("0"); continue; } if(k>65535) puts("ZenMeZheMeDuo"); else printf("%lld\n",k); } return 0; }
求出来ymax 再求出来ymin 再/a
非常简单的思路（虽然我觉得第一个更简单）
#include<bits/stdc++.h> using namespace std; #define ll long long #define A 100000 ll a,b,x,y,c,t; ll exgcd(ll a,ll b,ll &x,ll &y) { if(b==0) { x=1;y=0; return a; } ll c=exgcd(b,a%b,x,y); ll z=x; x=y;y=z-y*(a/b); // printf("x=%lld y=%lld\n",x,y); return c; } int main() { // freopen("data.in","r",stdin); // freopen("data.out","w",stdout); scanf("%lld",&t); while(t--) { scanf("%lld%lld%lld",&a,&b,&c); // printf("a=%lld b=%lld\n",a,b); x=0,y=0; if(a<0&&b<0) a=-a,b=-b,c=-c; if(a==0) { if((b<=0&&c>=0)||(b>=0&&c<=0)) { printf("0\n"); continue; } else if(c%b){ printf("0\n"); continue; } else{printf("ZenMeZheMeDuo\n");continue;}; } if(b==0) { if((a<=0&&c>=0)||(a>=0&&c<=0)) { printf("0\n"); continue; } else if(c%a){ printf("0\n"); continue; } else {printf("ZenMeZheMeDuo\n");continue;}; } ll g=exgcd(a,b,x,y),ans=0; x*=c/g,y*=c/g; // printf("%lld %lld a=%lld b=%lld c=%lld\n",x,y,a,b,c); if(c%g){printf("0\n");continue;} if(a*b<0) {printf("ZenMeZheMeDuo\n");continue;} a/=g,b/=g,c/=g;x%=b; while(x<=0) x+=b; y=(c-a*x)/b; ll y2=y%a; while(y2<=0) y2+=a; ans=(y-y2)/a+1; if(y2>y) ans=0; // printf("y=%lld y2=%lld x=%lld c=%lld\n",y,y2,x,c); if(ans<=65535) printf("%lld\n",ans); else printf("ZenMeZheMeDuo\n"); } }
题目思路还是挺简单的就是一些恶心的特判
首先$c \mod gcd！=0$时无解
然后$a，b$异号时无穷多解
$a$为$0$，$b$为$0$，$c$为$0$时无穷多解
$a$为$0$ $b$为$0$ $c$不为$0$ 无解
$ymax<ymin$无解
$a==0$ $ b,c$异号无解
$a==0$ $ b,c$同号无穷多解
$b==0$ $ a,c$异号无解
$b==0$ $ a,c$同号无穷多解
标签：typedef scan freopen else 恶心 exgcd include mes mat
原文地址：https://www.cnblogs.com/znsbc-13/p/11227871.html
