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

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概述

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

Sample Output


YES
YES
NO
NO

Source

BestCoder Round #12

#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

Sample Output


YES
NO

Source

BestCoder Round #12

#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


1 1 1
2 1 1

Sample Output


0.666667
0.187500

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

#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;
}