标签：adl 处理 sqrt type haoi2008 记忆 zoj repeat noip
又来写题解辣～然而并不太清楚题目排列情况。。。不管辣先写起来～
T1:［bzoj1041］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1041
一个赤果果的数学题～～～
首先坐标轴上一定有四个，
最大公约数搞一下然后判断一下就可以啦，细节全部都在代码～
代码如下：
1 var i,j,ans,d:longint; 2 t,r,m:int64; 3 function flag(x,y:longint):longint; 4 var t:longint; 5 begin 6 if (x<y) then 7 begin 8 t:=x; 9 x:=y; 10 y:=t; 11 end; 12 x:=x mod y; 13 if (x=0) then exit(y) else exit(flag(x,y)); 14 end; 15 begin 16 readln(r); 17 ans:=0; 18 t:=2*r; 19 for d:=1 to trunc(sqrt(t)) do 20 if (t mod d=0) then 21 begin 22 for i:=1 to trunc(sqrt(r/d)) do 23 begin 24 j:=trunc(sqrt(t/d-i*i)); 25 if (j=sqrt(t/d-i*i)) and (i<>j) and (flag(i,j)=1) then inc(ans); 26 end; 27 for i:=1 to trunc(sqrt(d/2)) do 28 begin 29 j:=trunc(sqrt(d-i*i)); 30 if (j*j=d-i*i) and (i<>j) and (flag(i,j)=1) then inc(ans); 31 end; 32 end; 33 ans:=ans*4+4; 34 writeln(ans); 35 end.
T2:［bzoj1042］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1042
又一个数学题hhh
先动态规划预处理，然后容斥原理就可以了。
代码如下：
1 var i,j,k,l,tot,g:longint; 2 c,s:array[1..4] of longint; 3 f:array[0..100010] of int64; 4 ans:int64; 5 function f1(x:longint):int64; 6 begin 7 if (x>=0) then exit(f[x]) else exit(0); 8 end; 9 begin 10 fillchar(c,sizeof(c),0); 11 fillchar(f,sizeof(f),0); 12 readln(c[1],c[2],c[3],c[4],tot); 13 f[0]:=1; 14 for i:=1 to 4 do 15 for j:=c[i] to 100000 do 16 f[j]:=f[j]+f[j-c[i]]; 17 for i:=1 to tot do 18 begin 19 fillchar(s,sizeof(s),0); 20 for j:=1 to 4 do 21 read(s[j]); 22 readln(g); 23 ans:=f[g]; 24 for j:=1 to 4 do 25 ans:=ans-f1(g-(s[j]+1)*c[j]); 26 for j:=1 to 3 do 27 for k:=j+1 to 4 do 28 if (j<>k) then 29 ans:=ans+f1(g-(s[j]+1)*c[j]-(s[k]+1)*c[k]); 30 for j:=1 to 2 do 31 for k:=j+1 to 3 do 32 for l:=k+1 to 4 do 33 if (j<>k) and (k<>l) and (j<>l) then 34 ans:=ans-f1(g-(s[j]+1)*c[j]-(s[k]+1)*c[k]-(s[l]+1)*c[l]); 35 ans:=ans+f1(g-(s[1]+1)*c[1]-(s[2]+1)*c[2]-(s[3]+1)*c[3]-(s[4]+1)*c[4]); 36 writeln(ans); 37 end; 38 end.
(T3先留好坑。。。之后补）
T4:［bzoj1044］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1044
第一问就是NOIP2015的day2T1（把我坑死的QAQQQ），
第二问就是动态规划就行了。
代码如下：
1 var n,m,i,j,k,ans,left,right,max,anss,s,now1,x,y:longint; 2 l,sl,ty:array[0..50000] of longint; 3 f,f1:array[0..1,-1..50000] of longint; 4 function tryit(x:longint):boolean; 5 var i,j,s,left,right:longint; 6 begin 7 i:=0; 8 j:=0; 9 now1:=0; 10 while (i<n) do 11 begin 12 left:=i+1; 13 right:=n; 14 i:=(left+right) div 2; 15 while (right-left>1) do 16 begin 17 if (sl[i]-sl[now1]<=x) then left:=i else right:=i; 18 i:=(left+right) div 2; 19 end; 20 if (sl[right]-sl[now1]<=x) then i:=right else i:=left; 21 if (sl[n]-sl[now1]>x) then inc(j); 22 now1:=i; 23 end; 24 if (j>m) then exit(false) else exit(true); 25 end; 26 begin 27 read(n,m); 28 fillchar(l,sizeof(l),0); 29 fillchar(sl,sizeof(sl),0); 30 s:=0; 31 max:=0; 32 for i:=1 to n do 33 begin 34 read(l[i]); 35 sl[i]:=sl[i-1]+l[i]; 36 if (l[i]>max) then max:=l[i]; 37 s:=s+l[i]; 38 end; 39 left:=max; 40 right:=s; 41 ans:=(left+right) div 2; 42 while (right-left>1) do 43 begin 44 if not(tryit(ans)) then left:=ans else right:=ans; 45 ans:=(left+right) div 2; 46 end; 47 ans:=right; 48 if tryit(left) then ans:=left; 49 write(ans,‘ ‘); 50 fillchar(f,sizeof(f),0); 51 fillchar(ty,sizeof(ty),0); 52 fillchar(f1,sizeof(f1),0); 53 anss:=0; 54 f[1,0]:=1; 55 for i:=0 to n do 56 f1[1,i]:=1; 57 k:=0; 58 for j:=1 to n do 59 begin 60 while (sl[j]-sl[k]>ans) do inc(k); 61 ty[j]:=k; 62 end; 63 for i:=1 to m+1 do 64 begin 65 x:=i mod 2; 66 y:=1-x; 67 for j:=0 to n do 68 begin 69 f[y,j]:=(f1[x,j-1]-f1[x,ty[j]-1]+10007) mod 10007; 70 f1[y,j]:=(f[y,j]+f1[y,j-1]) mod 10007; 71 end; 72 anss:=(anss+f[y,n]) mod 10007; 73 end; 74 writeln(anss); 75 end.
T5:［bzoj1045］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1045
这题。。。标准的组合数学嘛～～～
随便搞一下就粗来了。。。QAQ
辣么辣么多数学题QAQQQ
代码如下：
1 var n,i,j:longint; 2 sum,ave,ans:int64; 3 a,p:array[0..1000000] of longint; 4 procedure qsort(lx,rx:longint); 5 var i,j,m,x:longint; 6 begin 7 i:=lx; 8 j:=rx; 9 m:=p[(i+j) div 2]; 10 repeat 11 while (p[i]<m) do inc(i); 12 while (p[j]>m) do dec(j); 13 if (i<=j) then 14 begin 15 x:=p[i]; 16 p[i]:=p[j]; 17 p[j]:=x; 18 inc(i); 19 dec(j); 20 end; 21 until (i>j); 22 if (i<rx) then qsort(i,rx); 23 if (j>lx) then qsort(lx,j); 24 end; 25 begin 26 readln(n); 27 fillchar(a,sizeof(a),0); 28 fillchar(p,sizeof(p),0); 29 sum:=0; 30 for i:=1 to n do 31 begin 32 readln(a[i]); 33 sum:=sum+a[i]; 34 end; 35 ave:=sum div n; 36 for i:=2 to n do 37 p[i]:=p[i-1]-ave+a[i]; 38 qsort(1,n); 39 ans:=0; 40 for i:=1 to n do 41 ans:=ans+abs(p[i]-p[n div 2+1]); 42 writeln(ans); 43 end.
T6：［bzoj1054］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1054
QAQ这题是一个赤果果的BFS。。。
不对。。。
BFS要判重。。。这题判重都不要hhh
代码如下：
1 type arr=array[1..4,1..4] of longint; 2 const modp=10000009; 3 tx:array[1..4] of longint=(1,-1,0,0); 4 ty:array[1..4] of longint=(0,0,1,-1); 5 var i,j,k,s,t,s1,s2,rx,ry,swap,rh:longint; 6 q:array[0..65536*2] of arr; 7 mark:array[0..65536*2] of boolean; 8 step:array[0..65536*2] of longint; 9 ans:arr; 10 ch:char; 11 function hash(x:arr):longint; 12 var i,j,ans,now:longint; 13 begin 14 ans:=0; 15 now:=1; 16 for i:=1 to 4 do 17 for j:=1 to 4 do 18 begin 19 ans:=(ans+int64(now*x[i,j])); 20 now:=(now+now); 21 end; 22 exit(ans); 23 end; 24 begin 25 s:=0; 26 t:=1; 27 fillchar(q,sizeof(q),0); 28 fillchar(ans,sizeof(ans),0); 29 for i:=1 to 4 do 30 begin 31 for j:=1 to 4 do 32 begin 33 read(ch); 34 while (ch<>‘0‘) and (ch<>‘1‘) do read(ch); 35 q[0,i,j]:=ord(ch)-48; 36 end; 37 end; 38 for i:=1 to 4 do 39 begin 40 for j:=1 to 4 do 41 begin 42 read(ch); 43 while (ch<>‘0‘) and (ch<>‘1‘) do read(ch); 44 ans[i,j]:=ord(ch)-48; 45 end; 46 end; 47 s1:=hash(q[s]);s2:=hash(ans); 48 mark[s1]:=true; 49 fillchar(step,sizeof(step),0); 50 if (s1=s2) then writeln(‘0‘) else 51 begin 52 while (s<t) do 53 begin 54 for i:=1 to 4 do 55 for j:=1 to 4 do 56 if (q[s,i,j]>0) then 57 begin 58 for k:=1 to 4 do 59 begin 60 rx:=i+tx[k];ry:=j+ty[k]; 61 if (1<=rx) and (rx<=4) and (1<=ry) and (ry<=4) and (q[s,rx,ry]=0) then 62 begin 63 swap:=q[s,i,j];q[s,i,j]:=q[s,rx,ry];q[s,rx,ry]:=swap; 64 rh:=hash(q[s]); 65 if not(mark[rh]) then 66 begin 67 if (rh=s2) then 68 begin 69 writeln(step[s]+1); 70 halt; 71 end; 72 mark[rh]:=true; 73 q[t]:=q[s]; 74 step[t]:=step[s]+1; 75 inc(t); 76 end; 77 swap:=q[s,i,j];q[s,i,j]:=q[s,rx,ry];q[s,rx,ry]:=swap; 78 end; 79 end; 80 end; 81 inc(s); 82 end; 83 end; 84 end.
T7:［bzoj1055］
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1055
又一个动态规划呢！
不过是一个记忆化搜索。。。
代码如下：
1 const ch:array[1..4] of char=(‘W‘,‘I‘,‘N‘,‘G‘); 2 var i,j:longint; 3 f,flag:array[1..4,1..200,1..200] of boolean; 4 x:array[1..4,1..16,1..2] of char; 5 a:array[1..4] of longint; 6 s:string; 7 flagg:boolean; 8 function calc(x:char):longint; 9 begin 10 if (x=‘W‘) then exit(1) 11 else if (x=‘I‘) then exit(2) 12 else if (x=‘N‘) then exit(3) 13 else if (x=‘G‘) then exit(4); 14 end; 15 function f1(k,left,right:longint):boolean; 16 var i,j,t1,t2:longint; 17 begin 18 if flag[k,left,right] then exit(f[k,left,right]); 19 flag[k,left,right]:=true; 20 if (right=left) and (s[left]=ch[k]) then 21 begin 22 f[k,left,right]:=true; 23 exit(true); 24 end; 25 for i:=1 to a[k] do 26 begin 27 t1:=calc(x[k,i,1]); 28 t2:=calc(x[k,i,2]); 29 for j:=left to right-1 do 30 if f1(t1,left,j) and f1(t2,j+1,right) then 31 begin 32 f[k,left,right]:=true; 33 exit(true); 34 end; 35 end; 36 f[k,left,right]:=false; 37 exit(false); 38 end; 39 begin 40 fillchar(a,sizeof(a),0); 41 for i:=1 to 4 do 42 read(a[i]); 43 readln; 44 fillchar(f,sizeof(f),false); 45 fillchar(flag,sizeof(flag),false); 46 for j:=1 to 4 do 47 for i:=1 to a[j] do 48 readln(x[j,i,1],x[j,i,2]); 49 readln(s); 50 flagg:=false; 51 for i:=1 to 4 do 52 if f1(i,1,length(s)) then 53 begin 54 write(ch[i]); 55 flagg:=true; 56 end; 57 if not(flagg) then write(‘The name is wrong!‘); 58 writeln; 59 end.
（T8的坑也先留着。。。QAQQQ）
终于写完啦～～～撒花撒花～～～
标签：adl 处理 sqrt type haoi2008 记忆 zoj repeat noip
原文地址：http://www.cnblogs.com/Tommyr7/p/6055703.html
