这就没什么说的了，好好读题就行。
1 #include<cstring> 2 #include<cstdio> 3 #include<bits/stdc++.h> 4 using namespace std; 5 6 int main() 7 { 8 int n; 9 cin>>n; 10 int a1,b1,a2,b2; 11 int ok=5; 12 cin>>a1>>b1; 13 if(a1!=b1) ok=0; 14 for(int i=1;i<n;i++) 15 { 16 cin>>a2>>b2; 17 if(a2!=b2) ok=0; 18 if(ok==5&&a1<a2) ok=1; 19 a1=a2; 20 b1=b2; 21 } 22 if(ok==1) puts("unrated"); 23 if(ok==0) puts("rated"); 24 if(ok==5) puts("maybe"); 25 }
每一步转移时两种决策，加100或者减50
1 #include<cstring> 2 #include<cstdio> 3 #include<bits/stdc++.h> 4 using namespace std; 5 const int maxn=2000010; 6 int vis[maxn]; 7 int p,x,y; 8 int check(int n) 9 { 10 int i=(n/50)%475; 11 int ct=0; 12 for(;ct<25;ct++) 13 { 14 i=(i*96+42)%475; 15 if(i+26==p) return 1; 16 } 17 return 0; 18 } 19 20 struct node 21 { 22 int x,d; 23 bool operator<(const node &f)const 24 { 25 return d>f.d; 26 } 27 }now,nex; 28 29 int bfs() 30 { 31 int ans=1e9; 32 memset(vis,0,sizeof(vis)); 33 now.x=x; 34 now.d=0; 35 vis[x]=1; 36 priority_queue<node> q; 37 q.push(now); 38 while(!q.empty()) 39 { 40 now=q.top(); 41 q.pop(); 42 int xx=now.x; 43 if(check(xx)) return now.d; 44 if(xx-50>=y&&!vis[xx-50]) 45 { 46 nex.x=now.x-50; 47 nex.d=now.d; 48 vis[xx-50]=1; 49 q.push(nex); 50 } 51 if(!vis[xx+100]) 52 { 53 nex.x=now.x+100; 54 vis[xx+100]=1; 55 nex.d=now.d+1; 56 q.push(nex); 57 } 58 } 59 return ans; 60 } 61 62 int main() 63 { 64 65 cin>>p>>x>>y; 66 int ok=bfs(); 67 if(ok==1e9) printf("-1 "); 68 else printf("%d ",ok); 69 70 }
由题意可得，
(x+a)/(y+b)=p/q
x+a=k*p , y+b=k*q
a=k*p-x , b=k*q-y
我们要求的是b的最小值，显然b关于k单调，二分查找k的最小值即可。
（第一反应是扩展欧几里得-_-||）
1 #include <cstdio> 2 #include <cstring> 3 #define LL long long 4 5 int main() 6 { 7 int n; 8 int x,y,p,q; 9 scanf("%d",&n); 10 while(n--) 11 { 12 scanf("%d%d%d%d",&x,&y,&p,&q); 13 LL l=0,r=0x3f3f3f3f; 14 LL ans=-1; 15 while(l<=r) 16 { 17 LL m=(l+r)/2; 18 LL a=m*p-x; 19 LL b=m*q-y; 20 if(a>=0&&b>=0&&a<=b) 21 { 22 r=m-1; 23 ans=b; 24 } 25 else l=m+1; 26 } 27 printf("%lld ",ans); 28 } 29 }