109 条题解
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3
PowderHan LV 10 @ 2017-05-08 08:59:00
/* 水题啊 直接做就好了 Orz 果然是考数学的OTZ */ #include <iostream> #include <cstdio> #include <cmath> using namespace std; int n; int cnt; double H,S,V,L,K; int main () { cin>>H>>S>>V>>L>>K>>n; double tmax=sqrt(H/5),tmin=sqrt((H-K)/5); double x1=S-V*tmax,x2=S-V*tmin+L; for (int x=0;x<n;x++) if (x>=x1-0.00001&&x<=x2+0.00001) cnt++; cout<<cnt<<endl; return 0; } -
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/* 水题啊 直接做就好了 Orz 果然是考数学的OTZ */ #include <iostream> #include <cstdio> #include <cmath> using namespace std; int n; int cnt; double H,S,V,L,K; int main () { cin>>H>>S>>V>>L>>K>>n; double tmax=sqrt(H/5),tmin=sqrt((H-K)/5); double x1=S-V*tmax,x2=S-V*tmin+L; for (int x=0;x<n;x++) if (x>=x1-0.00001&&x<=x2+0.00001) cnt++; cout<<cnt<<endl; return 0; } -
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事实证明，数据规模这么小的情况下，直接每个都比较一遍何乐而不为呢？非要用一个式子算出来反而要思考具体怎么算以及特例，徒增自己的烦恼。
最后的max就是因为x_max经过min了以后，考虑一辆s1特别大的车可能导致x_min比x_max反而要大了，减出个负数个球。改完就对了。不正经
#include <iostream> #include <cmath> using namespace std; typedef long double ld; const ld ERR = 0.00001; const ld g = 10; ld h, s1, v, l, k, n, cnt; int main() { cin >> h >> s1 >> v >> l >> k >> n; ld t_ground_min = sqrt(2 * (h - k) / g); ld t_ground_max = sqrt(2 * h / g); // chops off unecessary bits. ld x_car_min = max(s1 - t_ground_max * v - ERR, (ld)-1); ld x_car_max = min(s1 - t_ground_min * v + l + ERR, n - 1 + ERR); // cout << x_car_min << " " << x_car_max << endl; cout << max((int)x_car_max - (int)x_car_min, 0) << endl; return 0; }正经
#include <iostream> #include <cmath> using namespace std; typedef long double ld; const ld ERR = 0.00001; const ld g = 10; ld h, s1, v, l, k, n, cnt; int main() { cin >> h >> s1 >> v >> l >> k >> n; ld t_ground_min = sqrt(2 * (h - k) / g); ld t_ground_max = sqrt(2 * h / g); // chops off unecessary bits. ld x_car_min = s1 - t_ground_max * v; ld x_car_max = s1 - t_ground_min * v + l; int cnt = 0; for (int x = 0; x <= n - 1; x++) { if (x + ERR > x_car_min && x - ERR < x_car_max) cnt++; } cout << cnt << endl; return 0; } -
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#include<cstdio> #include<iostream> #include<cmath> #define UNC_DIS 0.00001 using namespace std; int main(){ double h, s, v, l, k; int n, tot = 0; cin >> h >> s >> v >> l >> k >> n; for(int i=0; i<n; i++){ double tmin = (s-i*1.0)/v, tmax = (s+l-i*1.0)/v; double bpup = h - 5.0*tmin*tmin + UNC_DIS, bpdown = h - 5.0*tmax*tmax - UNC_DIS; if((bpup <= k && bpup >= 0.0) || (bpdown >= 0.0 && bpdown <= k) || (bpup >= k && bpdown <= 0.0)){ tot++; } } cout << tot; return 0; } -
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var h,s1,v,l,k,t1,t2:double; n,d1,d2,ans:longint; function min(a,b:longint):longint; begin if a>b then exit(b) else exit(a); end; function max(a,b:longint):longint; begin if a>b then exit(a) else exit(b); end; begin readln(h,s1,v,l,k,n); t1:=sqrt(0.2*(h-k)); //writeln(t1); t2:=sqrt(0.2*(h)); // writeln(t2); d1:=min(trunc(s1-v*t1+0.00001+l),n); // writeln(d1); d2:=max(trunc(s1-v*t2+0.00001),0); // writeln(d2); if d2=s1-v*t2+0.00001 then ans:=d1-d2+1 else ans:=d1-d2; writeln(ans); end.``` -
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function min(a,b:longint):longint; begin if a>b then exit(b) else exit(a); end; function max(a,b:longint):longint; begin if a>b then exit(a) else exit(b); end; begin readln(h,s1,v,l,k,n); t1:=sqrt(0.2*(h-k)); //writeln(t1); t2:=sqrt(0.2*(h)); // writeln(t2); d1:=min(trunc(s1-v*t1+0.00001+l),n); // writeln(d1); d2:=max(trunc(s1-v*t2+0.00001),0); // writeln(d2); if d2=s1-v*t2+0.00001 then ans:=d1-d2+1 else ans:=d1-d2; writeln(ans); end.``` -
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caesar_199 LV 9 @ 2016-09-04 08:16:20
#include<iostream> #include<cstdio> #include<cmath> using namespace std; int n; int cnt; double H, S, V, L, K; int main () { // freopen("in.txt", "r", stdin); cin >> H >> S >> V >> L >> K >> n; double tmax = sqrt(H/5), tmin = sqrt((H-K)/5); double x1 = S - V*tmax, x2 = S - V*tmin + L; for (int x = 0; x < n; x++) if (x >= x1-0.00001 && x <= x2+0.00001) cnt++; cout << cnt; return 0; } -
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sbxiaobai7 LV 8 @ 2016-02-20 18:00:51
接到球有两种情况：
1、落到车顶上
2、在球落到地前小车迎面撞上去#include<iostream>
#include<cmath>
using namespace std;
const double MIN = 1e-5;
const double g = 10.0;
int main()
{
int ans = 0;
double h, s1, v, l, k, n;
cin >> h >> s1 >> v >> l >> k >> n;
double t1 = sqrt(2 * (h - k) / g), t2 = sqrt(2 * h / g);
for(int i = n-1;i >= 0;i--) {
if((s1 - v * t1 - i) <= MIN && (s1 - v * t1 - i) >= -l - MIN)
ans++;
else if((s1 - v * t2 - i) < MIN && (s1 - v * t1 - i) > MIN)
ans++;
}
cout << ans;
} -
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qq872191552 LV 9 @ 2015-06-03 14:52:05
原来这道题里说的0.00001就是为了考虑左右车壁吸球的情况
一开始只有80分，于是怒把t1、t2减了0.00001，看能不能把左右车壁吸球转化成纵向接球，结果AC#include<stdio.h>
#include<math.h>
int main( )
{
float h,sl,v,l,k,t1,t2,g1,g2;
int ans=0,i,n;scanf("%f %f %f %f %f %d",&h,&sl,&v,&l,&k,&n);
for(i=n-1;i>=0;i--)
{
t1=(sl-i)/v-0.00001;
t2=(sl-i+l)/v-0.00001;g1=h-(5*t1*t1);
g2=h-(5*t2*t2);if(g1-k<=0.00001 && g1>=0) ans++;
else if(g1-k>0.00001 && g2-k<=0.00001) ans++;}
printf("%d",ans);
return 0;
} -
0
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
float H,S,V,L,K,n,t1,t2,d1,d2,tot=0;
cin>>H>>S>>V>>L>>K>>n;t1=sqrt(2*H/10);
t2=sqrt(2*(H-K)/10);
d1=S-V*t2+L;
d2=S-V*t1;
for(int i=0;i<n;++i)
if(i<=d1+0.00001&&i>=d2-0.00001)
tot=tot+1;
cout<<tot;
return 0;
} -
0
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>#define M 101001
#define INF 0x7fffffffusing namespace std;
double H,S1,V,L,K,f1,f2;
int n,num=0;
double a[M];
double tminx,tmaxx;void init()
{
scanf("%lf%lf%lf%lf%lf",&H,&S1,&V,&L,&K);
scanf("%d",&n);
for(int i=1;i<=n;i++)
a[i]=(i-1)*1.0;
tmaxx=sqrt(2.0*H/10.0);
tminx=sqrt(2.0*(H-K)/10.0);
f1=S1-tminx*V+L;
f2=S1-tmaxx*V;
for(int i=1;i<=n;i++)
{
if(a[i]>=f2-0.00001&&a[i]<=f1+0.00001)
num++;
}
printf("%d\n",num);
}int main()
{
init();
return 0;
}
根据小球高度在车子可以接受的范围内判断车子移动的距离能否接到小球 -
0
Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步 -
0
精炼13行，公式推导爆一切！！！！！！！！！！！！
program gh;
var h,s,v,l,k,t,t1,g:real;
n,i,total:longint;
begin
read(h,s,v,l,k,n);
g:=10;
t:=sqrt((h-k)*2/g);
t1:=sqrt(h*2/g);
total:=0;
for i:=n-1 downto 0 do
if ((s-v*t-i)/v<=t1-t+0.00001) and (s+l-v*t>=i) then inc(total);
write(total);
end. -
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Jinpeng1998 LV 7 @ 2012-08-08 16:51:29
VijosNT Mini 2.0.5.7 Special for Vijos
编译通过...
├ 测试数据 01：答案正确... (130ms, 580KB)
├ 测试数据 02：答案错误... (146ms, 580KB)
├ 测试数据 03：答案错误... (107ms, 580KB)
├ 测试数据 04：答案正确... (185ms, 580KB)
├ 测试数据 05：答案错误... (126ms, 580KB)---|---|---|---|---|---|---|---|-
Unaccepted / 40 / 315ms / 580KBview sourceprint?01 Var
02 h,s1,v,l,k,time,head,tail,zan:real;
03 n,ans:longint;
04 Begin
05 read(h,s1,v,l,k,n);
06 time:=sqrt((h-k)/5);
07 tail:=s1-time*v;
08 head:=tail+l;
09 if tail< 0 then tail:=0;
10 if head>N then head:=n;
11 ans:=trunc(head+0.00001)-trunc(tail+0.00001);
12 zan:=tail-trunc(tail);
13 if zan =0 then ans:=ans+1;
14 if ans
PowderHan
LV 10
@ 2017-05-08 08:59:00
