378 条题解
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0
福尔摩斯 LV 3 @ 2006-10-04 14:41:04
刚开始没有考虑到分差大于等于2时才算赢，结果只拿了50分。
后来加上这一步之后就AC了！ -
0
SharnHoiste LV 3 @ 2006-09-27 18:22:30
注意3点
1. 一般情况下 首先达到11分或者21分的会赢
2. 达到11分或21分后， 必须比对手多两分这局才算赢 不然一直比下去。
3. Puppy的服务器没有任何问题 和对于Fish等服务器有的时候第7组数据会出现答案错误 莫名其妙。 -
0
LV 0 @ 2006-09-11 19:57:40
其实就是边读边做，将结果记录到一个数组里，最后统一输出（别忘了输出当前比分，包括Star_Gym大牛提醒的0：0）。我这样做一次就A了。
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-1
wu767439815 LV 5 @ 2021-05-19 09:38:56
import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.List; import java.util.Scanner; /** * @author wpx * @version V1.0 * @Package com.algorithm * @date 2021/5/18 14:05 */ public class Main { public static void main(String[] args) { Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in))); StringBuilder builder = new StringBuilder(); while(true){ final String oneLine = sc.nextLine(); int endPos = oneLine.indexOf("E"); if(endPos != -1){ // 表明有结束符 builder.append(oneLine.substring(0, endPos).replaceAll("\\s", "")); break; } else { builder.append(oneLine.replaceAll("\\s", "")); } } String str = builder.toString(); int wNum = 0; int lNum = 0; List<String> resultOne = new ArrayList<>(); for(int i = 0; i < str.length(); i++){ String winner = String.valueOf(str.charAt(i)); if(winner.equals("W")){ wNum++; } else { lNum++; } // 判断是否已经有胜负 if(wNum >= 11 || lNum >= 11) { // 判断是否存在加赛的情况 if(wNum >= 10 && lNum >= 10) { // 判断比分差异是否大于2 if(wNum - lNum == 2 || wNum - lNum == -2){ resultOne.add(wNum + ":" + lNum); lNum = 0; wNum = 0; } } else { resultOne.add(wNum + ":" + lNum); lNum = 0; wNum = 0; } } } resultOne.add(wNum + ":" + lNum); lNum = 0; wNum = 0; List<String> resultTwo = new ArrayList<>(); for(int i = 0; i < str.length(); i++){ String winner = String.valueOf(str.charAt(i)); if(winner.equals("W")){ wNum++; } else { lNum++; } // 判断是否已经有胜负 if(wNum >= 21 || lNum >= 21) { // 判断是否存在加赛的情况 if(wNum >= 20 && lNum >= 20) { // 判断比分差异是否大于2 if(wNum - lNum == 2 || wNum - lNum == -2){ resultTwo.add(wNum + ":" + lNum); lNum = 0; wNum = 0; } } else { resultTwo.add(wNum + ":" + lNum); lNum = 0; wNum = 0; } } } resultTwo.add(wNum + ":" + lNum); for(String tmpObj : resultOne){ System.out.println(tmpObj); } System.out.println(); for(String tmpObj : resultTwo){ System.out.println(tmpObj); } } } -
-1
0:0的属实坑爹
match = [] list1 = [] s = 0 w = 0 l = 0 while True: list1 = list(input()) for i in list1: if i != 'E': match.append(i) else: s = 1 break if s == 1: break for i in match: if i == 'W': w+=1 if i == 'L': l+=1 if (w >= 11 or l >= 11) and (w - l >= 2 or l - w >= 2): print('{:.0f}:{:.0f}'.format(w,l)) w = 0 l = 0 continue if w > 0 or l > 0: print('{:.0f}:{:.0f}'.format(w,l)) if w == 0 and l == 0: print("0:0") print() w = 0 l = 0 for i in match: if i == 'W': w+=1 if i == 'L': l+=1 if (w >= 21 or l >= 21) and (w - l >= 2 or l - w >= 2): print('{:.0f}:{:.0f}'.format(w,l)) w = 0 l = 0 continue if w > 0 or l > 0: print('{:.0f}:{:.0f}'.format(w,l),end=' ') if w == 0 and l == 0: print("0:0",end=' ') -
-1
用python关键要注意,print是默认回车的!
坑了我1h啊这个问题
用print("23333", end='')可以取消默认会车Str = '' while 1: a = input() if 'E' in a: Str+=a.split("E")[0] break else: Str+=a #11分 nW=0;nL=0 for i in Str: if i == 'W': nW+=1 else: nL+=1 if (nW>=11 or nL>=11) and abs(nW-nL)>=2: print("%s:%s\n" % (nW,nL)) nW=0;nL=0 print("%s:%s\n" % (nW,nL)) print('\n') #21分 nW=0;nL=0 for i in Str: if i == 'W': nW+=1 else: nL+=1 if (nW>=21 or nL>=21) and abs(nW-nL)>=2: print("%s:%s\n" % (nW,nL)) nW=0;nL=0 print("%s:%s" % (nW,nL), end='') -
-1
#include<stdio.h>
#include<string.h>
int main()
{
char s[100000]={0},c;
int i=0,W=0,L=0,x=0,y=0,cnt;
scanf("%c",&c);
while(c!='E')
{
s[i++]=c;
if(c=='W')W++;
else if (c=='L')L++;
if((W-L>=2&&W>=11)||(L-W>=2&&L>=11))
{
printf("%d:%d\n",W,L);
L=0,W=0;
}
scanf("%c",&c);
}
printf("%d:%d\n\n",W,L);
for(cnt=0;cnt<i;cnt++)
{
if(s[cnt]=='W')x++;
else if (s[cnt]=='L')y++;
if((x-y>=2&&x>=21)||(y-x>=2&&y>=21))
{
printf("%d:%d\n",x,y);
x=0,y=0;
}
}
printf("%d:%d\n\n",x,y);
return 0;
} -
-1
#include<stdio.h>
#include<string.h>
int main()
{
char s[100000]={0},c;
int i=0,W=0,L=0,x=0,y=0,cnt;
scanf("%c",&c);
while(c!='E')
{
s[i++]=c;
if(c=='W')W++;
else if (c=='L')L++;
if((W-L>=2&&W>=11)||(L-W>=2&&L>=11))
{
printf("%d:%d\n",W,L);
L=0,W=0;
}
scanf("%c",&c);
}
printf("%d:%d\n\n",W,L);
for(cnt=0;cnt<i;cnt++)
{
if(s[cnt]=='W')x++;
else if (s[cnt]=='L')y++;
if((x-y>=2&&x>=21)||(y-x>=2&&y>=21))
{
printf("%d:%d\n",x,y);
x=0,y=0;
}
}
printf("%d:%d\n\n",x,y);
return 0;
} -
-1
#include <iostream>
using namespace std;
int main()
{
char wl;
int sa11=0,sb11=0,sa21=0,sb21=0,s21[1000000][2],i;
for(i=0;i<1000000;i++) {s21[i][0]=0;s21[i][1]=0;}
i=0;
do
{
cin>>wl;
if(wl=='W') {sa11++;sa21++;}
else if(wl=='L') {sb11++;sb21++;}
if((sa11>=11&&sa11-sb11>=2)||(sb11>=11&&sb11-sa11>=2))
{
cout<<sa11<<':'<<sb11<<endl;
sa11=0;sb11=0;
}
if((sa21>=21&&sa21-sb21>=2)||(sb21>=21&&sb21-sa21>=2))
{
s21[i][0]=sa21;s21[i][1]=sb21;i++;
sa21=0;sb21=0;
}
}
while(wl!='E');
if(s21[0][0]==0&&s21[0][1]==0&&sa21==0&&sb21==0)
cout<<"0:0\n\n0:0";
if(sa21!=0||sb21!=0) {s21[i][0]=sa21;s21[i][1]=sb21;}
else i--;
if((s21[0][0]!=0||s21[0][1]!=0)&&sa11==0&&sb11==0)
cout<<"0:0\n";
if(sa11!=0||sb11!=0) cout<<sa11<<':'<<sb11<<endl;
cout<<endl;
for(int j=0;j<=i;j++)
cout<<s21[j][0]<<':'<<s21[j][1]<<endl;
return 0;
} -
-1
#include <iostream>
using namespace std;
int main(void) {
char A[100000];
int left = 0, right = 0;
int i = 0;
while ((A[i] = getchar()) != EOF) {
if (A[i] == 'E')break;
++i;
}
i = 0;
while (A[i] != 'E') {
if (A[i] == 'W')
++left;
else if (A[i] == 'L')
++right;
if ((left >= 11 || right >= 11) && (left - right > 1 || right - left > 1)) {
cout << left <<':'<< right << endl;
left = right = 0;
}
++i;
}
cout << left << ':' << right << endl;
cout << endl;
left = right = i = 0;
while (A[i] != 'E') {
if (A[i] == 'W')
++left;
else if (A[i] == 'L')
++right;
if ((left >= 21 || right >= 21) && (left - right > 1 || right - left > 1)) {
cout << left << ':' << right << endl;
left = right = 0;
}
++i;
}
cout << left << ':' << right;
return 0;
} -
-1
#include <iostream> #include <queue> using namespace std; class Result { public: int playerA; int playerB; int max; Result(int max) { this->max = max; playerA = 0; playerB = 0; } void add(char c) { if (c == 'W') { playerA++; } else { playerB++; } } bool isEnd() { return (playerA >= max && playerA - playerB > 1) || (playerB >= max && playerB - playerA > 1); } }; class Recorder { public: queue<Result> records; int max; Recorder(int max) { this->max = max; } void add(char c) { prepareQueue(); records.back().add(c); } void prepareQueue() { if (records.empty() || records.back().isEnd()) { Result r(max); records.push(r); } } }; ostream &operator<<(ostream &os, Result &r) { os << r.playerA << ":" << r.playerB; return os; } ostream &operator<<(ostream &os, Recorder &r) { if (r.records.empty()) { os << "0:0" << endl; } while (!r.records.empty()) { os << r.records.front() << endl; r.records.pop(); } return os; }; int main() { Recorder r11(11); Recorder r21(21); while (true) { char c; cin >> c; if (c == 'E' ) { break; } if (c == 'W' || c == 'L') { r11.add(c); r21.add(c); } } // bug of the quiz: if previous game ends, it expect we have one more 0:0 r11.prepareQueue(); r21.prepareQueue(); cout << r11 << endl << r21; return 0; } -
-1
#include<iostream>
#include<string>
#include<math.h>
using namespace std;
int abs(int a){
if(a<0) a=0-a;
return a;
}
int main(){
char a[100001];
int w,l,i;
w=0;
l=0;
i=1;
while(cin>>a[i]&&a[i]!='E'){
i+=1;
}
i=i-1;
for(int k=1;k<=i;k++){
if(a[k]=='W') w+=1;
if(a[k]=='L') l+=1;
if(w>=11||l>=11){
if(abs(w-l)>=2){
cout<<w<<":"<<l<<endl;
w=0;
l=0;
}
}
}cout<<w<<":"<<l<<endl<<endl;
w=0;
l=0;for(int k=1;k<=i;k++){
if(a[k]=='W') w+=1;
if(a[k]=='L') l+=1;
if(w>=21||l>=21){
if(abs(w-l)>=2){
cout<<w<<":"<<l<<endl;
w=0;
l=0;
}
}
}cout<<w<<":"<<l<<endl<<endl;
}
福尔摩斯
LV 3
@ 2006-10-04 14:41:04
