BC12haoj5058&&hdoj5059&&hdoj5560 So easy Help him War

113 阅读 0 评论 75 点赞

我是靠谱客的博主辛勤美女，这篇文章主要介绍BC12haoj5058&&hdoj5059&&hdoj5560 So easy Help him War，现在分享给大家，希望可以做个参考。

So easy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 755 Accepted Submission(s): 402

Problem Description

Small W gets two files. There are n integers in each file. Small W wants to know whether these two files are same. So he invites you to write a program to check whether these two files are same. Small W thinks that two files are same when they have the same integer set.
For example file A contains (5,3,7,7),and file B contains (7,5,3,3). They have the same integer set (3,5,7), so they are same.
Another sample file C contains(2,5,2,5), and file D contains (2,5,2,3).
The integer set of C is (2,5),but the integer set of D is (2,3,5),so they are not same.
Now you are expected to write a program to compare two files with size of n.

Input

Multi test cases (about 100). Each case contain three lines. The first line contains one integer n represents the size of file. The second line contains n integers

a1,a2,a3,…,an - represents the content of the first file. The third line contains n integers

b1,b2,b3,…,bn - represents the content of the second file.
Process to the end of file.

1≤n≤100

1≤ai,bi≤1000000000

Output

For each case, output "YES" (without quote) if these two files are same, otherwise output "NO" (without quote).

Sample Input

复制代码

Sample Output

复制代码

1
2
3
4
5
6

YES
YES
NO
NO

Source

BestCoder Round #12

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#include<map>
using namespace std;
const int maxn=110;
map<int,int>visa,visb;
int numa[maxn];
int numb[maxn];
int main()
{
int t,i,j,k,n,m;
while(scanf("%d",&n)!=EOF){
visa.clear();
visb.clear();
int a,b;
for(i=0;i<n;++i){
scanf("%d",&numa[i]);
visa[numa[i]]++;
}
for(i=0;i<n;++i){
scanf("%d",&numb[i]);
visb[numb[i]]++;
}
for(i=0;i<n;++i){
if(!visa.count(numb[i])||!visb.count(numa[i]))break;
}
if(i<n)
printf("NOn");
else
printf("YESn");
}
return 0;
}

Help him

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2194 Accepted Submission(s): 445

Problem Description

As you know, when you want to hack someone's program, you must submit your test data. However sometimes you will submit invalid data, so we need a data checker to check your data. Now small W has prepared a problem for BC, but he is too busy to write the data checker. Please help him to write a data check which judges whether the input is an integer ranged from a to b (inclusive).
Note: a string represents a valid integer when it follows below rules.
1. When it represents a non-negative integer, it contains only digits without leading zeros.
2. When it represents a negative integer, it contains exact one negative sign ('-') followed by digits without leading zeros and there are no characters before '-'.
3. Otherwise it is not a valid integer.

Input

Multi test cases (about 100), every case occupies two lines, the first line contain a string which represents the input string, then second line contains a and b separated by space. Process to the end of file.

Length of string is no more than 100.
The string may contain any characters other than 'n','r'.
-1000000000

≤a≤b≤1000000000

Output

For each case output "YES" (without quote) when the string is an integer ranged from a to b, otherwise output "NO" (without quote).

Sample Input

复制代码

Sample Output

复制代码

1
2
3
4

YES
NO

Source

BestCoder Round #12

复制代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
char str[110];
char stra[110];
char strb[110];
int main()
{
int i,j,k;
while(gets(str)){
scanf("%s%s",stra,strb);
getchar();
int len=strlen(str);
bool sign=true;
for(i=0;i<len;++i){
if(str[i]=='-'&&i==0)continue;
if(sign&&str[i]=='0')break;
if(str[i]>='0'&&str[i]<='9');
else break;
sign=false;
}
if((len==1&&str[0]=='0')){
i=len+1;sign=false;
}
if(i<len||sign){
printf("NOn");
}
else {
if(str[0]=='-'){
if(stra[0]=='-'&&strb[0]=='-'){
if(strcmp(stra+1,str+1)>=0&&strcmp(strb+1,str+1)<=0)
printf("YESn");
else
printf("NOn");
}
else if(stra[0]=='-'&&strb[0]!='-'){
if(strcmp(stra+1,str+1)>=0)
printf("YESn");
else
printf("NOn");
}
else {
printf("NOn");
}
}
else {
if(stra[0]=='-'&&strb[0]=='-'){
printf("NOn");
}
else if(stra[0]=='-'&&strb[0]!='-'){
if(strcmp(strb,str)>=0)
printf("YESn");
else
printf("NOn");
}
else {
if(strcmp(stra,str)<=0&&strcmp(strb,str)>=0)
printf("YESn");
else
printf("NOn");
}
}
}
}
return 0;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
char str[110];
char stra[110];
char strb[110];
int main()
{
int i,j,k;
while(gets(str)){
scanf("%s%s",stra,strb);
getchar();
int len=strlen(str);
bool sign=true;
for(i=0;i<len;++i){
if(str[i]=='-'&&i==0)continue;
if(sign&&str[i]=='0')break;
if(str[i]>='0'&&str[i]<='9');
else break;
sign=false;
}
if((len==1&&str[0]=='0')){
i=len+1;sign=false;
}
if(i<len||sign){
printf("NOn");
}
else {
if(str[0]=='-'){
if(stra[0]=='-'&&strb[0]=='-'){
if(strcmp(stra+1,str+1)>=0&&strcmp(strb+1,str+1)<=0)
printf("YESn");
else
printf("NOn");
}
else if(stra[0]=='-'&&strb[0]!='-'){
if(strcmp(stra+1,str+1)>=0)
printf("YESn");
else
printf("NOn");
}
else {
printf("NOn");
}
}
else {
if(stra[0]=='-'&&strb[0]=='-'){
printf("NOn");
}
else if(stra[0]=='-'&&strb[0]!='-'){
if(strcmp(strb,str)>=0)
printf("YESn");
else
printf("NOn");
}
else {
if(strcmp(stra,str)<=0&&strcmp(strb,str)>=0)
printf("YESn");
else
printf("NOn");
}
}
}
}
return 0;
}

War

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 233 Accepted Submission(s): 76
Special Judge

Problem Description

Long long ago there are two countrys in the universe. Each country haves its own manor in 3-dimension space. Country A's manor occupys x^2+y^2+z^2<=R^2. Country B's manor occupys x^2+y^2<=HR^2 && |z|<=HZ. There may be a war between them. The occurrence of a war have a certain probability.
We calculate the probability as follow steps.
1. VC=volume of insection manor of A and B.
2. VU=volume of union manor of A and B.
3. probability=VC/VU

Input

Multi test cases(about 1000000). Each case contain one line. The first line contains three integers R,HR,HZ. Process to end of file.

[Technical Specification]
0< R,HR,HZ<=100

Output

For each case，output the probability of the war which happens between A and B. The answer should accurate to six decimal places.

Sample Input

复制代码

Sample Output

复制代码

Source

BestCoder Round #12

题意：求圆柱体与球相交的体积和相并的体积的积

解题思路分情况讨论：球冠的体积计算公式V = πh*h（r-h/3） r为原球的半径h为球冠的高

复制代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#define PI acos(-1.0)
using namespace std;
const int maxn=10010;
int main()
{
double r,hr,hz;
while(scanf("%lf%lf%lf",&r,&hr,&hz)!=EOF){
double ans;
if(hr>r){
if(hz>r){
double hc=4.0/3.0*PI*r*r*r;
double hu=PI*hr*hr*hz*2.0;
ans=hc/hu;
}
else {
double v=PI*(r-hz)*(r-hz)*(r-(r-hz)/3.0)*2.0;
double hc=4.0/3.0*PI*r*r*r-v;
double hu=PI*hr*hr*hz*2.0+4.0/3.0*PI*r*r*r-hc;
ans=hc/hu;
}
}
else {
if(hz<=sqrt(r*r-hr*hr)){
double hc=PI*hr*hr*hz*2.0;
double hu=4.0/3.0*PI*r*r*r;
ans=hc/hu;
}
else if(hz>r){
double hh=sqrt(r*r-hr*hr);
double hc=PI*hr*hr*hh*2.0+PI*(r-hh)*(r-hh)*(r-(r-hh)/3.0)*2.0;
double hu=4.0/3.0*PI*r*r*r+PI*hr*hr*hz*2.0-hc;
ans=hc/hu;
}
else {
double hh=sqrt(r*r-hr*hr);
double v1=PI*(r-hh)*(r-hh)*(r-(r-hh)/3.0);
double v2=PI*(r-hz)*(r-hz)*(r-(r-hz)/3.0);
double hc=PI*hr*hr*hh*2.0+(v1-v2)*2.0;
double hu=4.0/3.0*PI*r*r*r+PI*hr*hr*hz*2.0-hc;
ans=hc/hu;
}
}
printf("%.6lfn",ans);
}
return 0;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#define PI acos(-1.0)
using namespace std;
const int maxn=10010;
int main()
{
double r,hr,hz;
while(scanf("%lf%lf%lf",&r,&hr,&hz)!=EOF){
double ans;
if(hr>r){
if(hz>r){
double hc=4.0/3.0*PI*r*r*r;
double hu=PI*hr*hr*hz*2.0;
ans=hc/hu;
}
else {
double v=PI*(r-hz)*(r-hz)*(r-(r-hz)/3.0)*2.0;
double hc=4.0/3.0*PI*r*r*r-v;
double hu=PI*hr*hr*hz*2.0+4.0/3.0*PI*r*r*r-hc;
ans=hc/hu;
}
}
else {
if(hz<=sqrt(r*r-hr*hr)){
double hc=PI*hr*hr*hz*2.0;
double hu=4.0/3.0*PI*r*r*r;
ans=hc/hu;
}
else if(hz>r){
double hh=sqrt(r*r-hr*hr);
double hc=PI*hr*hr*hh*2.0+PI*(r-hh)*(r-hh)*(r-(r-hh)/3.0)*2.0;
double hu=4.0/3.0*PI*r*r*r+PI*hr*hr*hz*2.0-hc;
ans=hc/hu;
}
else {
double hh=sqrt(r*r-hr*hr);
double v1=PI*(r-hh)*(r-hh)*(r-(r-hh)/3.0);
double v2=PI*(r-hz)*(r-hz)*(r-(r-hz)/3.0);
double hc=PI*hr*hr*hh*2.0+(v1-v2)*2.0;
double hu=4.0/3.0*PI*r*r*r+PI*hr*hr*hz*2.0-hc;
ans=hc/hu;
}
}
printf("%.6lfn",ans);
}
return 0;
}