381 条题解
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-1
白浅星零 LV 7 @ 2019-06-17 13:36:41
#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;
}
-
-1
感觉异常繁琐，又不知道怎么改...
#include <stdio.h>int main()
{
int i=0,huahua=0,duishou=0,state=1;
char n[2];
int game[100000]={0};
scanf("%1s",n);
while (n[0]!='E')
{
if(n[0]=='W')
{
game[i]=1;
}else if (n[0]=='L')
{
game[i]=2;
}
i++;
scanf("%1s",n);
}
i=0;
if (game[0]==0)
{
printf("0:0\n\n0:0");
}else
{
while (game[i]!=0)
{
huahua=0;
duishou=0;
while ((((huahua<11)&&(duishou<11))||((huahua-duishou<2)&&(duishou-huahua<2)))&&(game[i]!=0))
{
if (game[i]==1)
{
huahua++;
}else
{
duishou++;
}
i++;
}
printf("%d:%d\n",huahua,duishou);
}
if ((huahua+duishou)%11==0)
{
printf("0:0\n");
}
printf("\n");
i=0;
while (game[i]!=0)
{
huahua=0;
duishou=0;
while ((((huahua<21)&&(duishou<21))||((huahua-duishou<2)&&(duishou-huahua<2)))&&(game[i]!=0))
{
if (game[i]==1)
{
huahua++;
}else
{
duishou++;
}
i++;
}
printf("%d:%d\n",huahua,duishou);
}
if ((huahua+duishou)%21==0)
{
printf("0:0");
}
}return 0;
} -
-1
livingstone LV 6 @ 2019-02-09 12:30:42
好像有点繁琐了……
#include <iostream> #include <cmath> using namespace std; int main(){ int PointHua=0,PointDui=0,i=0,j=0,k=0; char rec[100000],cha; while (cin>>rec[i]) i++; while (cha=rec[j]){ j++; if (cha=='W') PointHua++; if (cha=='L') PointDui++; if (cha=='E') break; if (PointHua>=11||PointDui>=11){ if (abs(PointHua-PointDui)>1){ cout<<PointHua<<':'<<PointDui<<endl; PointHua=0; PointDui=0; } } } cout<<PointHua<<':'<<PointDui<<endl; cout<<endl; PointHua=0; PointDui=0; while (cha=rec[k]){ k++; if (cha=='W') PointHua++; if (cha=='L') PointDui++; if (cha=='E') break; if (PointHua>=21||PointDui>=21){ if (abs(PointHua-PointDui)>1){ cout<<PointHua<<':'<<PointDui<<endl; PointHua=0; PointDui=0; } } } cout<<PointHua<<':'<<PointDui<<endl; return 0; } -
-1
#Python 很繁琐，请多指教 rawscorelist = list(input()) scorelist = [] while 'E' not in rawscorelist: scorelist.extend(rawscorelist) rawscorelist = list(input()) for i in range(len(rawscorelist)): if rawscorelist[i] == 'E': break else: scorelist.append(rawscorelist[i]) def test(A): X = 0 Y = 0 n = 0 for i in scorelist: if i == "W": X = X + 1 if i == 'L': Y = Y + 1 if abs(X - Y) >= 2 and (X >= A or Y >= A): print("{}:{}".format(X, Y)) X = Y = 0 print("{}:{}".format(X, Y)) test(11) print() test(21) -
-1
#include<iostream>
#include<cmath>
using namespace std;
int main()
{ char jilu[100000];
for(int i=0;;i++)
{
cin>>jilu[i];
if(jilu[i]=='E')
break;}
int x,y;
x=0;y=0;
for(int i=0;;i++)
{
if(jilu[i]=='W')
x++;
if(jilu[i]=='L')
y++;if((x>=11||y>=11)&&abs(x-y)>=2)
{
cout<<x<<":"<<y<<endl;
x=0;y=0;
}
if(jilu[i]=='E')
{
cout<<x<<":"<<y<<endl;
break;
}}
x=0;y=0;
cout<<endl;
for(int i=0;;i++)
{
if(jilu[i]=='W')
x++;
if(jilu[i]=='L')
y++;if((x>=21||y>=21)&&abs(x-y)>=2)
{
cout<<x<<":"<<y<<endl;
x=0;y=0;
}
if(jilu[i]=='E')
{
cout<<x<<":"<<y<<endl;
break;
}}return 0;
} -
-1
Johnwegner LV 6 @ 2018-08-13 23:56:30
加了注释的解决方案 适合C++
#include<iostream>
#include<string>
#include<cstring>using std::endl;
using std::cin;
using std::cout;
using std::string;void process(string, int);
int main()
{
string resultline[100000]; //开1000个空间过不了，hhh
string result;
int nail = 0;
while (1)
{
getline(cin, resultline[nail]);
if (resultline[nail] == "") break;
nail += 1;
} //可能读到有E的，后面再处理
for (int i = 0; i <= nail; i++)
{
result += resultline[i];
}process(result, 11);
cout<<endl;
process(result, 21);return 0;
}void process(string result, int s) //按不同标准进行处理
{
int i = 0;// 作为计数
int hua = 0; //华华分数
int huaop = 0; //华华对手分数
while (1)
{
if (result[i] == 'W') hua += 1;
if (result[i] == 'L') huaop += 1;
if ( result[i] == 'E' ) break;
i += 1;
if ((((hua == s) || (huaop == s)) && (abs(hua - huaop) >= 2)) || (((hua>s)||(huaop>s)) && (abs(hua-huaop)==2))) //未加赛超两分，加赛直接超过两分进行下一局
{
cout << hua << ":" << huaop << endl;
hua = huaop = 0;
}
}
cout << hua << ":" << huaop << endl; // 最后一场没打满，但比分还在
} -
-1
哈哈 (hjw1879541824) LV 6 @ 2018-07-14 07:21:25
#include<cstdio>
using namespace std;
char c[100000001];int main()
{
int i(0);
while(scanf("%c",&c[++i])&&c[i]!='E');
int m(0),n(0);
for(int j=1;j<=i;++j)
{
if(c[j]=='W')++m;
if(c[j]=='L')++n;
if(m+n==11||j==i)
{
printf("%d:%d\n",m,n);
m=0;n=0;
}
}
m=0;n=0;
printf("\n");
for(int j=1;j<=i;++j)
{
if(c[j]=='W')++m;
if(c[j]=='L')++n;
if(m+n==21||j==i)
{
printf("%d:%d\n",m,n);
m=0;n=0;
}
}return 0;
}//哪里有错？样例全过。
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-1
wy_doubixiaoxin LV 7 @ 2018-07-11 15:26:37
if name == '__main__':
# with open("Demo3_test", 'r')as file:
# string = file.read()
string = ''
while 1:
s = input()
string += s
if s.find('E') != -1:
break
W_count = 0
L_count = 0
for i in string:
if i == 'E':
print(str(W_count) + ':' + str(L_count))
break
else:
if i == 'W':
W_count += 1
if i == 'L':
L_count += 1
if (W_count >= 11 or L_count >= 11) and (abs(W_count - L_count) > 1):
print(str(W_count) + ':' + str(L_count))
W_count = 0
L_count = 0
print()
W_count = 0
L_count = 0
for i in string:
if i == 'E':
print(str(W_count) + ':' + str(L_count))
break
else:
if i == 'W':
W_count += 1
if i == 'L':
L_count += 1
if (W_count >= 21 or L_count >= 21) and (abs(W_count-L_count) > 1):
print(str(W_count) + ':' + str(L_count))
W_count = 0
L_count = 0-
yangjz1125 @ 2018-08-25 21:01:51
python 没缩进！！！ game over
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-
-1
#include <stdio.h> #include <stdlib.h> #include <limits.h> #define MAXLEN 100000 int main() { char ch; int ele[MAXLEN][2], twel[MAXLEN][2]; int a = 0, b = 0; while ((ch = getchar()) != 'E') { if (ch == '\n' || ch == ' ') continue; ele[a][ch=='L']++; twel[b][ch=='L']++; if ((ele[a][0] >= 11 || ele[a][1] >= 11) && abs(ele[a][0]-ele[a][1]) >= 2) a++; if ((twel[b][0] >= 21 || twel[b][1] >= 21) && abs(twel[b][0]-twel[b][1]) >= 2) b++; } for (int i = 0; i <= a; i++) { printf("%d:%d\n", ele[i][0], ele[i][1]); } printf("\n"); for (int j = 0; j <= b; j++) { printf("%d:%d\n", twel[j][0], twel[j][1]); } return 0; } -
-1
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms用最原始的方法做，一直TLE，。。。。搞了半天数组开小了
开到10000迅速ac
。。。。。。。。。真是BT啊，，浪费我4次提交
-
-1
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
怕被封号..............程序代码删了.............. -
-1
编译通过...
├ 测试数据 01：运行超时|无输出...
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案错误... ├ 标准行输出
├ 错误行输出
---|---|---|---|---|---|---|---|-
Unaccepted 有效得分：80 有效耗时：0ms这个是怎么回事？ 标准行输出怎么会是0：0啊？不可能吧！！再我第一个怎么超时
白浅星零
LV 7
@ 2019-06-17 13:36:41
