A.数三角形
题目：http://www.contesthunter.org/contest/CH%20Round%20%2348%20-%20Streaming%20%233%20(NOIP模拟赛Day1)/数三角形
题解：暴力枚举三元组，判断是否共线即可，用叉积
代码：
1 type cp=record 2 x,y:double; 3 end; 4 var a:array[0..200] of cp; 5 i,j,k,n,ans:longint; 6 function cross(a,b,c:cp):double; 7 begin 8 cross:=(b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y); 9 end; 10 procedure init; 11 begin 12 readln(n); 13 for i:=1 to n do readln(a[i].x,a[i].y); 14 end; 15 procedure main; 16 begin 17 ans:=0; 18 for i:=1 to n do 19 for j:=i+1 to n do 20 for k:=j+1 to n do 21 begin 22 if cross(a[i],a[j],a[k])<>0 then inc(ans); 23 end; 24 writeln(ans); 25 end; 26 27 begin 28 init; 29 main; 30 end.
B.4和7
题目：http://www.contesthunter.org/contest/CH%20Round%20%2348%20-%20Streaming%20%233%20(NOIP模拟赛Day1)/4和7
题解：这类似于vijosp1002过河，那题是当距离超过105时缩到105，但我取20也过了，现在也不知道为什么要压。。。
这一道题比较好做，因为只有4和7两个数，由裴蜀定理我们知道 （4，7）=1 所以仅有4和7是可以组合出所有的整数的，
但这题必须保证4和7的系数都必须是非负的。
所以我们从前面开始尝试，看看能发现什么规律。
发现前面有的可以4和7组合出来 比如15，但有的就组合不出来，如13
进一步我们又发现当x>=18时，就都可以组合出来了。我们详细的证明一下：
i） x mod 4=0 直接全用4即可
ii) x mod 4=1 我们可以用3个7，剩下的全用4，因为这样的数最小是21
iii）x mod 4=2 我们用2个7，剩下的全用4
iiii）x mod 4=3 我们用1个7，剩下的全用4
这样就可以了当相邻两个有药的位置距离相隔≥18时就将它看作18，这样是没有问题的。
考场上没有处理刚开始的边界情况，WA了30分
代码：
1 var v,a,b,c,f:array[-100..2000000] of int64; 2 i,n:longint; 3 ans,m,tmp:int64; 4 function max(x,y:int64):int64; 5 begin 6 if x>y then exit(x) else exit(y); 7 end; 8 procedure sort(l,r:longint); 9 var i,j,x,y:longint; 10 begin 11 i:=l;j:=r;x:=b[(i+j)>>1]; 12 repeat 13 while b[i]<x do inc(i); 14 while b[j]>x do dec(j); 15 if i<=j then 16 begin 17 y:=a[i];a[i]:=a[j];a[j]:=y; 18 y:=b[i];b[i]:=b[j];b[j]:=y; 19 inc(i);dec(j); 20 end; 21 until i>j; 22 if i<r then sort(i,r); 23 if j>l then sort(l,j); 24 end; 25 procedure init; 26 begin 27 readln(n,m); 28 for i:=1 to n do readln(a[i],b[i]); 29 sort(1,n); 30 end; 31 procedure main; 32 begin 33 b[0]:=0;c[0]:=0; 34 for i:=1 to n do 35 if b[i]-b[i-1]>=18 then c[i]:=c[i-1]+18 36 else c[i]:=c[i-1]+b[i]-b[i-1]; 37 fillchar(v,sizeof(v),0); 38 for i:=1 to n do inc(v[c[i]],a[i]); 39 f[0]:=1; 40 for i:=4 to c[n] do 41 begin 42 tmp:=max(f[i-4],f[i-7]); 43 if tmp=0 then continue; 44 f[i]:=tmp+v[i]; 45 end; 46 ans:=0; 47 for i:=0 to c[n] do ans:=max(ans,f[i]); 48 writeln(ans-1); 49 end; 50 51 begin 52 init; 53 main; 54 end.
C.反射镜
题目：http://www.contesthunter.org/contest/CH%20Round%20%2348%20-%20Streaming%20%233%20(NOIP模拟赛Day1)/反射镜
题解：考场上没想到什么好的算法，于是打了200行的暴力。。。幸亏没写跪，骗到60分，结果标程竟然那么短。。。
代码：
1.暴力
1 type cp=record 2 x,y,id,idx:longint; 3 z:char; 4 end; 5 var a,b:array[0..150000] of cp; 6 now,dir,n,m,i,x,y,tmp:longint; 7 c:array[0..150000] of longint; 8 dist,t:int64; 9 ans:cp; 10 11 function cmp1(a,b:cp):boolean; 12 begin 13 if a.x=b.x then exit(a.y<b.y) else exit(a.x<b.x); 14 end; 15 function cmp2(a,b:cp):boolean; 16 begin 17 if a.y=b.y then exit(a.x<b.x) else exit(a.y<b.y); 18 end; 19 procedure sort1(l,r:longint); 20 var i,j:longint;x,y:cp; 21 begin 22 i:=l;j:=r;x:=a[(i+j)>>1]; 23 repeat 24 while cmp1(a[i],x) do inc(i); 25 while cmp1(x,a[j]) do dec(j); 26 if i<=j then 27 begin 28 y:=a[i];a[i]:=a[j];a[j]:=y; 29 inc(i);dec(j); 30 end; 31 until i>j; 32 if i<r then sort1(i,r); 33 if j>l then sort1(l,j); 34 end; 35 procedure sort2(l,r:longint); 36 var i,j:longint;x,y:cp; 37 begin 38 i:=l;j:=r;x:=b[(i+j)>>1]; 39 repeat 40 while cmp2(b[i],x) do inc(i); 41 while cmp2(x,b[j]) do dec(j); 42 if i<=j then 43 begin 44 y:=b[i];b[i]:=b[j];b[j]:=y; 45 inc(i);dec(j); 46 end; 47 until i>j; 48 if i<r then sort2(i,r); 49 if j>l then sort2(l,j); 50 end; 51 procedure init; 52 begin 53 readln(n,m,t); 54 for i:=1 to n do begin readln(a[i].x,a[i].y,a[i].z,a[i].z);a[i].id:=i;end; 55 for i:=1 to n do b[i]:=a[i]; 56 sort1(1,n); 57 sort2(1,n); 58 for i:=1 to n do c[a[i].id]:=i; 59 for i:=1 to n do b[i].idx:=c[b[i].id]; 60 for i:=1 to n do c[b[i].id]:=i; 61 for i:=1 to n do a[i].idx:=c[a[i].id]; 62 a[n+1].x:=maxlongint;a[n+1].y:=maxlongint; 63 a[0].x:=maxlongint;a[0].y:=maxlongint; 64 b[n+1]:=a[n+1];b[0]:=a[0]; 65 end; 66 procedure main; 67 begin 68 x:=0;y:=0; 69 for i:=1 to n do if (b[i].y=0) and (b[i].x>0) then begin now:=i;break;end; 70 if now=0 then 71 begin 72 writeln(t,' ',0);exit; 73 end; 74 dir:=3;dist:=b[now].x;ans:=b[now]; 75 while true do 76 begin 77 if dir=3 then 78 begin 79 if b[now].z='/' then 80 begin 81 dir:=2; 82 now:=b[now].idx; 83 tmp:=now+1; 84 if a[tmp].x<>a[now].x then break; 85 if dist+abs(a[tmp].y-a[now].y)>t then break 86 else inc(dist,abs(a[tmp].y-a[now].y)); 87 now:=tmp;ans:=a[now]; 88 end 89 else 90 begin 91 dir:=4; 92 now:=b[now].idx; 93 tmp:=now-1; 94 if a[tmp].x<>a[tmp].x then break; 95 if dist+abs(a[tmp].y-a[now].y)>t then break 96 else inc(dist,abs(a[tmp].y-a[now].y)); 97 now:=tmp;ans:=a[now]; 98 end; 99 end; 100 if dir=2 then 101 begin 102 if a[now].z='/' then 103 begin 104 dir:=3; 105 now:=a[now].idx; 106 tmp:=now+1; 107 if b[tmp].y<>b[now].y then break; 108 if dist+abs(b[tmp].x-b[now].x)>t then break 109 else inc(dist,abs(b[tmp].x-b[now].x)); 110 now:=tmp;ans:=b[now]; 111 end 112 else 113 begin 114 dir:=1; 115 now:=a[now].idx; 116 tmp:=now-1; 117 if b[tmp].y<>b[now].y then break; 118 if dist+abs(b[tmp].x-b[now].x)>t then break 119 else inc(dist,abs(b[tmp].x-b[now].x)); 120 now:=tmp;ans:=b[now]; 121 end; 122 end; 123 if dir=1 then 124 begin 125 if b[now].z='/' then 126 begin 127 dir:=4; 128 now:=b[now].idx; 129 tmp:=now-1; 130 if a[tmp].x<>a[now].x then break; 131 if dist+abs(a[tmp].y-a[now].y)>t then break 132 else inc(dist,abs(a[tmp].y-a[now].y)); 133 now:=tmp;ans:=a[now]; 134 end 135 else 136 begin 137 dir:=2; 138 now:=b[now].idx; 139 tmp:=now+1; 140 if a[tmp].x<>a[now].x then break; 141 if dist+abs(a[tmp].y-a[now].y)>t then break 142 else inc(dist,abs(a[tmp].y-a[now].y)); 143 now:=tmp;ans:=a[now]; 144 end; 145 end; 146 if dir=4 then 147 begin 148 if a[now].z='/' then 149 begin 150 dir:=1; 151 now:=a[now].idx; 152 tmp:=now-1; 153 if b[tmp].y<>b[now].y then break; 154 if dist+abs(b[tmp].x-b[now].x)>t then break 155 else inc(dist,abs(b[tmp].x-b[now].x)); 156 now:=tmp;ans:=b[now]; 157 end 158 else 159 begin 160 dir:=3; 161 now:=a[now].idx; 162 tmp:=now+1; 163 if b[tmp].y<>b[now].y then break; 164 if dist+abs(b[tmp].x-b[now].x)>t then break 165 else inc(dist,abs(b[tmp].x-b[now].x)); 166 now:=tmp;ans:=b[now]; 167 end; 168 end; 169 end; 170 if dir=1 then writeln(ans.x-(t-dist),' ',ans.y); 171 if dir=2 then writeln(ans.x,' ',ans.y+(t-dist)); 172 if dir=3 then writeln(ans.x+(t-dist),' ',ans.y); 173 if dir=4 then writeln(ans.x,' ',ans.y-(t-dist)); 174 end; 175 procedure spj; 176 begin 177 if a[1].y<>0 then writeln(t,' ',0) 178 else if a[1].x<0 then writeln(t,' ',0) 179 else if a[1].z='/' then writeln(a[1].x,' ',t-a[1].x) 180 else writeln(a[1].x,' ',a[1].x-t); 181 end; 182 183 begin 184 init; 185 if n=1 then spj else main; 186 end.
2.正解
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define UU 0 5 #define LL 1 6 #define DD 2 7 #define RR 3 8 //LeftTurn = 1 9 //RightTurn = 3 10 const int dx[4] = {0, -1, 0, 1}; 11 const int dy[4] = {1, 0, -1, 0}; 12 #define f(x, y, z) for(int x = (y); x <= (z); ++x) 13 #define f2(x, y, z) for(int x = (y), asdf = (z); x <= asdf; ++x) 14 using std::min; 15 using std::abs; 16 17 int N; long long T; 18 struct mir{int x, y; long long dr;} xm[400086], ym[400086]; 19 inline bool byx(const mir &a, const mir &b){ 20 return a.x < b.x || (a.x == b.x && a.y < b.y); 21 } 22 inline bool byy(const mir &a, const mir &b){ 23 return a.y < b.y || (a.y == b.y && a.x < b.x); 24 } 25 inline bool operator ==(const mir &a, const mir &b){ 26 return a.x == b.x && a.y == b.y; 27 } 28 29 inline mir *findxless(const mir a){ 30 int l = 0, r = N + 1; 31 while(l < r){ 32 int m = ((l + r) >> 1) + 1; 33 if(xm[m] == a) return xm + m; 34 else if(byx(xm[m], a)) l = m; else r = m - 1; 35 } 36 if(l < 1 || l > N || xm[l].x != a.x) return NULL; 37 return xm + l; 38 } 39 inline mir *findxmore(const mir a){ 40 int l = 0, r = N + 1; 41 while(l < r){ 42 int m = (l + r) >> 1; 43 if(xm[m] == a) return xm + m; 44 else if(byx(a, xm[m])) r = m; else l = m + 1; 45 } 46 if(l < 1 || l > N || xm[l].x != a.x) return NULL; 47 return xm + l; 48 } 49 inline mir *findyless(const mir a){ 50 int l = 0, r = N + 1; 51 while(l < r){ 52 int m = ((l + r) >> 1) + 1; 53 if(ym[m] == a) return ym + m; 54 else if(byy(ym[m], a)) l = m; else r = m - 1; 55 } 56 if(l < 1 || l > N || ym[l].y != a.y) return NULL; 57 return ym + l; 58 } 59 inline mir *findymore(const mir a){ 60 int l = 0, r = N + 1; 61 while(l < r){ 62 int m = (l + r) >> 1; 63 if(ym[m] == a) return ym + m; 64 else if(byy(a, ym[m])) r = m; else l = m + 1; 65 } 66 if(l < 1 || l > N || ym[l].y != a.y) return NULL; 67 return ym + l; 68 } 69 70 int main(){ 71 scanf("%d%*d%I64d", &N, &T); long long iT = T; 72 f(i, 1, N){ 73 int cx, cy; char buf[4]; 74 scanf("%d%d%s", &cx, &cy, buf); 75 xm[i].x = ym[i].x = cx; 76 xm[i].y = ym[i].y = cy; 77 if(buf[0] == '/'){ 78 xm[i].dr = 3; ym[i].dr = 1; 79 }else{ 80 xm[i].dr = 1; ym[i].dr = 3; 81 } 82 } 83 xm[0] = (mir) {-1000000008, -1000000008, 0}; 84 ym[0] = (mir) {-1000000008, -1000000008, 0}; 85 xm[N + 1] = (mir) {1000000008, 1000000008, 0}; 86 ym[N + 1] = (mir) {1000000008, 1000000008, 0}; 87 std::sort(xm + 1, xm + N + 1, byx); 88 std::sort(ym + 1, ym + N + 1, byy); 89 int cx = 0, cy = 0, cr = RR; 90 for(;;){ 91 // printf("C %d %d = %d ", cx, cy, cr); 92 mir *res; 93 if(cr == UU) res = findxmore((mir) {cx, cy + 1}); 94 else if(cr == LL) res = findyless((mir) {cx - 1, cy}); 95 else if(cr == DD) res = findxless((mir) {cx, cy - 1}); 96 else res = findymore((mir) {cx + 1, cy}); 97 if(res){ 98 if(cy == 0 && res->y == 0 && cx < 0 && res->x > 0) T %= (iT - T - cx); 99 int dis = abs(res->x - cx) + abs(res->y - cy); 100 if(dis < T){ 101 cx = res->x; cy = res->y; cr = (cr + res->dr) % 4; T -= (long long) dis; 102 }else{ 103 printf("%I64d %I64d ", (long long) cx + T * (long long) dx[cr], (long long) cy + T * (long long) dy[cr]); 104 return 0; 105 } 106 }else{ 107 printf("%I64d %I64d ", (long long) cx + T * (long long) dx[cr], (long long) cy + T * (long long) dy[cr]); 108 return 0; 109 } 110 } 111 return 0; 112 }