Codeforces 787
A
Problem description
有两串数b,b+a,b+2a....和d,d+c,d+2c..... 若两串数中会出现相同的数，则输出一个最小的相同的数，否则输出0
Data Limit：a,b,c,d <= 100 Time Limit: 1s
Solution
易得xa+b=yc+d=那个相同的数(x,y为自然数) 所以x=(yc+d-b)/a 又因为x为自然数，所以yc+d-b为a的倍数，即（yc+d-b）mod a为0 因为a小于100，所以余数的情况只有100种。那么y从0开始枚举，直到x成为整数。如果100次内x还无法成为整数，则打出-1. 再仔细想想，x和y都不会超过100，那两个for搞定。
Code
var a,b,c,d,i,l,t,ans,j:longint; function min(a,b:longint):longint; begin if a<b then exit(a) else exit(b); end; begin read(a,b); read(c,d); ans:=maxlongint; for i:=0 to 100 do for j:=0 to 100 do if i*a+b=c*j+d then ans:=min(ans,i*a+b); if ans<maxlongint then writeln(ans) else writeln(-1); end.
B
Problem description
给出n,m和m组数若m组数中都有至少一对相反数，就打NO，否则打YES
Data Limit：n，m <= 10000 Time Limit: 1s
Solution
每组暴力搜一下就好了，有一组不合题意，打出YES，然后halt
Code
var n,m,i,j,k,hh:longint; x:array[1..100000] of longint; a:array[1..10000000,1..3] of longint; bo:boolean; procedure jia(x:longint); var i,t:longint; begin t:=abs(x); for i:=1 to hh do if a[i,3]=t then begin if x>0 then inc(a[i,1]) else inc(a[i,2]); if (a[i,1]>0)and(a[i,2]>0) then begin bo:=true;end; exit; end; inc(hh); a[hh,3]:=t; if x>0 then inc(a[hh,1]) else inc(a[hh,2]); end; begin read(n,m); bo:=true; for i:=1 to m do begin if not bo then begin write('YES');halt;end; bo:=false; hh:=0; read(k); for j:=1 to k do read(x[j]); for j:=1 to n do begin a[j,1]:=0;a[j,2]:=0;a[j,3]:=0;end; for j:=1 to k do begin jia(x[j]); end; end; if not bo then writeln('YES') else writeln('NO'); end.
C
Problem description
两个人，每人有两组数，每人可以让怪兽前进自己数字的格数，谁先把怪兽移到1号位，就胜利 输出怪兽在每个位置时，两人分别先手时谁能赢
Data Limit：n <= 7000 Time Limit: 4s
Solution
设状态dp[x,y]表示当前怪物在x点，轮到玩家y操作的游戏结果。所以dp[1,0]和dp[1,1]的结果都是失败，若在dp[x,y]的情况下，玩家的任意操作到达的状态是y’玩家false，则dp[x,y]为true（玩家选择该操作即可胜利），反之，若玩家的所有选择到达的下一个状态都是另一个玩家y’胜利,则dp[x,y]为false. 不过因为会有loop的结果，所以递推不大好实现，我选择了使用逆向bfs的方法，有点像拓扑排序（先初始化所有点的度数为可操作数），先让点[0][0]和[0][1]入队，对于任意当前队列头的点，如果这个点的状态是false，则逆推出能从哪些点到这一点，那些点的状态都是true并且入队；如果这个点的状态是true，则让所有逆推出来的上一步的点的度数减一，如果该点的度数为0，则该点的状态是false并且入队。最后那些始终没有入队过的就是loop的点了。
Code
var n,i,j,h,t:longint; k0,k1:longint; a,b:array[1..100000] of longint; dp:array[1..100000,0..1] of boolean; q:array[1..1000000,1..2] of longint; d:array[1..100000,0..1] of longint; q3:array[1..1000000] of boolean; bo:array[1..1000000,0..1] of boolean; function s(a,b:longint):Longint; begin if a>b then exit(a-b); if a=b then exit(n); if a<b then exit(n+a-b); end; procedure push(x,y:longint;z:boolean); begin inc(h); q[h,1]:=x;q[h,2]:=y;q3[h]:=z;bo[x,y]:=true; dp[x,y]:=z; end; begin read(n); read(k0);for i:=1 to k0 do read(a[i]); read(k1);for i:=1 to k1 do read(b[i]); for i:=1 to n do begin d[i,0]:=k0;d[i,1]:=k1;end; push(1,0,false);push(1,1,false); repeat inc(t); if q[t,2]=0 then if not q3[t] then begin for i:=1 to k1 do if(not bo[s(q[t,1],b[i]),1]) then push(s(q[t,1],b[i]),1,true); end else begin for i:=1 to k1 do begin dec(d[s(q[t,1],b[i]),1]); if (d[s(q[t,1],b[i]),1]=0)and(not bo[s(q[t,1],b[i]),1]) then push(s(q[t,1],b[i]),1,false);end; end; if q[t,2]=1 then if not q3[t] then begin for i:=1 to k0 do if(not bo[s(q[t,1],a[i]),0]) then push(s(q[t,1],a[i]),0,true); end else begin for i:=1 to k0 do begin dec(d[s(q[t,1],a[i]),0]); if (d[s(q[t,1],a[i]),0]=0)and(not bo[s(q[t,1],a[i]),0]) then push(s(q[t,1],a[i]),0,false);end; end; until h<=t; for i:=2 to n do begin if (not bo[i,0]) then write('Loop ') else if dp[i,0] then write('Win ') else write('Lose '); end; writeln; for i:=2 to n do begin if not bo[i,1] then write('Loop ') else if dp[i,1] then write('Win ') else write('Lose '); end; end.
D
Problem description
给出一个有向图,求从起始点到各点的最小距离 有三种边: 1.从x到y的边 2.从x到l至r的边 3.从l至r到y的边
Data Limit：1 ≤ n, q ≤ 105, 1 ≤ s ≤ n Time Limit: 2s
Solution
暴力连边会超时,又因为是区间连边,所以用线段树 但要建两颗树,一颗从各点连到树上,另一颗从树上连到各点,然后做SPFA即可. 注:特判n=1的情况,数组不要开太小.
Code
var n,m,s,q,x,u,v,w,k,h,t,l,r,maxn:int64; i,j:longint; a,c,first,next,last,f,tree,intree,qq,arr:array[-1000000..1000000] of int64; q1:array[1..10000000] of int64; b:array[-1000000..1000000] of boolean; function max(a,b:int64):int64; begin if a>b then exit(a) else exit(b); end; procedure add(x,y,z:int64); begin //writeln(x,' ',y,' ',z); inc(k); a[k]:=y;;c[k]:=z; if first[x]=0 then first[x]:=k else next[last[x]]:=k; last[x]:=k; end; procedure build(root:int64;start,ed:int64); var mid:longint; begin if start=ed then begin tree[root]:=arr[ed];intree[arr[ed]]:=root;exit;end else begin mid:=(start+ed) shr 1; build(root*2,start,mid); build(root*2+1,mid+1,ed); end; end; procedure push(x:int64); begin inc(h); qq[h]:=x; end; procedure push1(x:int64); begin inc(t); q1[t]:=x; end; procedure add2(root,nstart,nend,ustart,uend,w,hhh:int64); var mid:longint; begin if (ustart>nend)or(uend<nstart) then exit; if (ustart<=nstart)and(nend<=uend) then begin add(hhh,root,w);// else add(hhh,-root,w); exit; end; mid:=(nstart+nend) shr 1; if root*2<=maxn then add2(root*2,nstart,mid,ustart,uend,w,hhh); if root*2+1<=maxn then add2(root*2+1,mid+1,nend,ustart,uend,w,hhh); end; procedure add3(root,nstart,nend,ustart,uend,w,hhh:int64); var mid:longint; begin if (ustart>nend)or(uend<nstart) then exit; if (ustart<=nstart)and(nend<=uend) then begin if tree[root]<>0 then add(root,hhh,w) else add(-root,hhh,w); exit; end; mid:=(nstart+nend) shr 1; if root*2<=maxn then add3(root*2,nstart,mid,ustart,uend,w,hhh); if root*2+1<=maxn then add3(root*2+1,mid+1,nend,ustart,uend,w,hhh); end; begin read(n,q,s); if n=1 then begin writeln(0); halt; end; for i:=1 to n do arr[i]:=i; build(1,1,n); push(1); repeat inc(t); add(qq[t],qq[t]*2+1,0); if tree[qq[t]*2+1]<>0 then add(qq[t]*2+1,-qq[t],0) else add(-(qq[t]*2+1),-qq[t],0); add(qq[t],qq[t]*2,0); if tree[qq[t]*2]<>0 then add(qq[t]*2,-qq[t],0) else add(-qq[t]*2,-qq[t],0); maxn:=max(maxn,qq[t]*2+1); if tree[qq[t]*2+1]=0 then begin push(qq[t]*2+1); end; if tree[qq[t]*2]=0 then begin push(qq[t]*2); end; until h<=t; for i:=1 to q do begin read(x); if x=1 then begin read(u,v,w);add(intree[u],intree[v],w);end; if x=2 then begin read(v,l,r,w); add2(1,1,n,l,r,w,intree[v]); end; if x=3 then begin read(v,l,r,w); add3(1,1,n,l,r,w,intree[v]); end; end; for i:=-maxn to maxn do f[i]:=20000000000; f[intree[s]]:=0; h:=0;t:=0; push1(intree[s]); repeat inc(h); b[q1[h]]:=false; x:=first[q1[h]]; while x<>0 do begin if f[q1[h]]+c[x]<f[a[x]] then begin f[a[x]]:=f[q1[h]]+c[x]; //writeln('f[',a[x],']=',f[a[x]],' ',x); if not b[a[x]] then push1(a[x]); end; x:=next[x]; end; until h>=t; for i:=1 to n do if f[intree[i]]<>20000000000 then write(f[intree[i]],' ') else write('-1 '); end.
