381 条题解
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0
蛟川洋 LV 5 @ 2018-03-25 18:41:13
建议p党练手就算了，毕竟字符串在c++里比pascal里恶心多了，思路详见代码。
#include<bits/stdc++.h> using namespace std ; //Vijos P1217 int ball ; string x , y ;//虽然我并不喜欢用string，但是这里的确用string更方便 void score(int x)//一个过程记录得分，表示在x分制下的结果 { int w = 0 , l = 0 ; for (int i = 0 ; i < y.size() ; i ++)//不知道哪里学来的一个测字符串长度的神奇函数 { if (y[i] == 'W')//赢球给自己加分 w ++ ; else//否则给对方加分，如题意 l ++ ; if (((w >= x) || (l >= x)) && (abs(w - l) >= 2))//按照乒乓球赛的规则，若满足一方达到分数，**同时**双方比方差距在2分以上即结束比赛[https://baike.baidu.com/item/%E4%B9%92%E4%B9%93%E7%90%83/221415?fr=aladdin](http://) { cout << w << ":" << l << endl ;//输出结果 w = 0 ; l = 0 ; } } cout << w << ":" << l << endl ;//最后有没比完的，也输出 } int main() { do { cin >> x ; ball = x.find("E") ;//也是一个神奇的函数，类似pascal中的pos if (ball == -1)//在c++中，未查到返回-1 y = y + x ; else { for (int i = 0 ; i < ball ; i ++) { y = y + x[i] ; } } } while (ball == -1) ; score(11) ;//调用score函数 cout << endl ; score(21) ; } -
0
聪明的猪 (xxz) LV 6 @ 2018-03-11 16:54:24
C++实现：
#include <iostream> #include <string> using std::string; void processing(string, int); int main(void) { using std::cin; using std::cout; using std::endl; bool run; int i; string result, winAndLose; run = true; while (run) { getline(cin, winAndLose); int len = int(winAndLose.length()); if(len > 20) { len = 20; } for (i = 0; i < len; i++) { switch (winAndLose.at(i)) { case '\0': run = false; i = int(winAndLose.length()); break; case 'E': run = false; i = int(winAndLose.length()); break; case 'W': result += winAndLose.at(i); break; case 'L': result += winAndLose.at(i); break; default: break; } } } processing(result, 11); cout << endl; processing(result, 21); return 0; } void processing(string winOrLose, int winScore) { using std::cout; using std::endl; int huaWin, huaLose, i; huaLose = 0; huaWin = 0; for(i = 0; i < int(winOrLose.length()); i++) { if ('W' == winOrLose.at(i)) { huaWin++; } else if ('L' == winOrLose.at(i)) { huaLose++; } if (((huaLose >= winScore) || (huaWin >= winScore)) && ((huaLose - huaWin > 1) || (huaWin - huaLose > 1))) { cout << huaWin << ":" << huaLose << endl; huaLose = 0; huaWin = 0; } } cout << huaWin << ":" << huaLose << endl; return; } -
0
C++实现：
#include <iostream> #include <string> using std::string; void processing(string, int); int main(void) { using std::cin; using std::cout; using std::endl; bool run; int i; string result, winAndLose; run = true; while (run) { getline(cin, winAndLose); int len = int(winAndLose.length()); if(len > 20) { len = 20; } for (i = 0; i < len; i++) { switch (winAndLose.at(i)) { case '\0': run = false; i = int(winAndLose.length()); break; case 'E': run = false; i = int(winAndLose.length()); break; case 'W': result += winAndLose.at(i); break; case 'L': result += winAndLose.at(i); break; default: break; } } } processing(result, 11); cout << endl; processing(result, 21); return 0; } void processing(string winOrLose, int winScore) { using std::cout; using std::endl; int huaWin, huaLose, i; huaLose = 0; huaWin = 0; for(i = 0; i < int(winOrLose.length()); i++) { if ('W' == winOrLose.at(i)) { huaWin++; } else if ('L' == winOrLose.at(i)) { huaLose++; } if (((huaLose >= winScore) || (huaWin >= winScore)) && ((huaLose - huaWin > 1) || (huaWin - huaLose > 1))) { cout << huaWin << ":" << huaLose << endl; huaLose = 0; huaWin = 0; } } cout << huaWin << ":" << huaLose << endl; return; } -
0
Java 版的
import java.io.IOException;
import java.util.Scanner;public class Main {
public static void main(String[] args) throws IOException{
Scanner sn = new Scanner(System.in);
String inputStr = "";
StringBuilder sb = new StringBuilder();
while(true){
String temp = sn.nextLine();
if (temp.contains("E")) {
sb.append(temp);
break;
}
sb.append(temp);
}
inputStr = sb.toString().substring(0,sb.toString().indexOf('E'));
countPoints(inputStr,11);
countPoints(inputStr,21);
}
public static void countPoints(String inputStr,int scoreType){
int score1 = 0;
int score2 = 0;
for(int i = 0;i<inputStr.length();i++){
if(inputStr.charAt(i) == 'W'){
score1 ++;
}
if(inputStr.charAt(i) == 'L'){
score2 ++;
}
if((score1 >=scoreType || score2 >= scoreType) && Math.abs(score1 -score2) >=2){
System.out.println(score1+":"+score2);
score1 = 0; score2 = 0;
}
}
System.out.println(score1+":"+score2);
System.out.println();
}
} -
0
#include<cstdio> using namespace std; int f21[10010][2]; int main(){ int w1=0,l1=0,w2=0,l2=0,j=0; char k; while(scanf("%c",&k)&&k!='E'){ if(k=='W'){w1++;w2++;} if(k=='L'){l1++;l2++;} if((w1>=11&&w1-l1>=2)||(l1>=11&&l1-w1>=2)){ printf("%d:%d\n",w1,l1); w1=0;l1=0; }; if((w2>=21&&w2-l2>=2)||(l2>=21&&l2-w2>=2)){ f21[j][0]=w2; f21[j][1]=l2; w2=0;l2=0;j++; }; } printf("%d:%d\n\n",w1,l1); for(int i=0;i<j;i++) printf("%d:%d\n",f21[i][0],f21[i][1]); printf("%d:%d",w2,l2); return 0; } -
0
1014763456 LV 5 @ 2018-02-08 17:09:13
#include <stdio.h>
int main()
{
char letter[1000];
int i,j;
int m=0,n=0;
int an=0,bn=0;
int a[10],b[10];
int en=0,fn=0;
int e[10],f[10];
for(i=0;i<1000;i++)
{
scanf("%c",&letter[i]);
if(letter[i]=='W')
{
an++;
en++;
}
if(letter[i]=='L')
{
bn++;
fn++;
}
if(an+bn==11)
{
a[m]=an;
b[m]=bn;
m++;
an=0;
bn=0;
}
if(en+fn==21)
{
e[n]=en;
f[n]=fn;
n++;
en=0;
fn=0;
}
if(letter[i]=='E')
break;
}
if(an!=0||bn!=0)
{
a[m]=an;
b[m]=bn;
}
if(en!=0||fn!=0)
{
e[n]=en;
f[n]=fn;
}
for(i=0;i<=m;i++)
printf("%d:%d\n",a[i],b[i]);
for(i=0;i<=n;i++)
printf("%d:%d\n",e[i],f[i]);
}大佬这个哪里错了 -
0
lijie201602 LV 6 @ 2018-01-21 14:39:43
#include<cstdio>
char s[1000010];
int a=0,b=0,n=0;
int main()
{
while(scanf("%c",&s[n]) && s[n]!='E')
if(s[n]!='\n') n++;
for(int i=0;i<n;i++)
{
if(s[i]=='W') a++;
if(s[i]=='L') b++;
if((a-b>1 || b-a>1) && (a>10 || b>10))
{
printf("%d:%d\n",a,b);
a=b=0;
}
}
printf("%d:%d\n",a,b);
printf("\n");
a=b=0;
for(int i=0;i<n;i++)
{
if(s[i]=='W') a++;
if(s[i]=='L') b++;
if((a-b>1 || b-a>1) && (a>20 || b>20))
{
printf("%d:%d\n",a,b);
a=b=0;
}
}
printf("%d:%d\n",a,b);
return 0;
} -
0
知道比赛规则的细节
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>int main()
{
char a[100000][22];
int i=0,j=0,w=0,l=0,n;
while((a[i][j]=getchar())!='E')
{
if(a[i][j]=='\n')
{
i++;j=0;
}
else j++;
}
n=i;
for(i=0;i<=n&&a[i-1][j]!='E';i++)
{
for(j=0;a[i][j]!='\n';j++)
{if(a[i][j]=='E')
break;
if(a[i][j]=='W')
w++;
if(a[i][j]=='L')
l++;
if((w==11||l==11)&&abs(w-l)>=2)
{
printf("%d:%d\n",w,l);
w=0;
l=0;
}
if((w>=10&&l>=10)&&abs(w-l)>=2)
{
printf("%d:%d\n",w,l);
w=l=0;
}
}
}
printf("%d:%d\n",w,l);
printf("\n");
w=l=0;
for(i=0;i<=n&&a[i-1][j]!='E';i++)
{
for(j=0;a[i][j]!='\n';j++)
{
if(a[i][j]=='E')
break;
if(a[i][j]=='W')
w++;
if(a[i][j]=='L')
l++;
if((l==21||w==21)&&abs(w-l)>=2)
{
printf("%d:%d\n",w,l);
w=0;
l=0;
}
if((w>=20&&l>=20)&&abs(w-l)>=2)
{
printf("%d:%d\n",w,l);
w=l=0;
}}
}
printf("%d:%d",w,l);
return 0;
} -
0
plzletmego LV 5 @ 2018-01-05 15:41:27
#include<iostream>
using namespace std;
#include<cstring>
char str[10000000];
int main()
{
for(int i=0;;++i)
{
cin>>str[i];
if(str[i]=='E') break;
}
int a=0,b=0,len=strlen(str);
for(int i=0;i!=len-1;++i)
{
if(str[i]==' ') continue;
if(str[i]=='W') ++a;
else ++b;
if((a>=11&&a-b>=2)||(b>=11&&b-a>=2)) {cout<<a<<":"<<b<<endl;a=0;b=0;}
}
cout<<a<<":"<<b<<endl<<endl;
a=0;
b=0;
for(int i=0;i!=len-1;++i)
{
if(str[i]==' ') continue;
if(str[i]=='W') ++a;
else ++b;
if((a>=21&&a-b>=2)||(b>=21&&b-a>=2)) {cout<<a<<":"<<b<<endl;a=0;b=0;}
}
cout<<a<<":"<<b<<endl;
} -
0
#include<stdio.h>
#include<math.h>//WWWWWWWWWWWWWWWWWWWWWWLWE
char ch[10010];
void outpute(int n,int type)
{
int i,a=0,b=0;
for(i=0;i<n;i++)
{if(ch[i]=='W') a++;
if(ch[i]=='L') b++;
if((a==type||b==type)&&(abs(a-b)>=2)||(abs(a-b)==2)&&((a>type)||(b>type)))
{
printf("%d:%d\n",a,b);
a=0,b=0;
}
}
printf("%d:%d",a,b);
}
int main()
{
int i=0;
do//----------------------方法2
{
scanf("%c",&ch[i]);
}while(ch[i++]!='E');
/* while ((ch[i] = getchar()) != EOF) {
if (ch[i] == 'E') break;
i++;
}*/
outpute(i,11);
printf("\n\n");
outpute(i,21);
return 0;
} -
0
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
int xiaohua=0;
int enemy=0;
string str;
string::iterator it;
getline(cin,str,'E');
it=str.begin();
if(str=="")
{
cout<<"0:0"<<endl;
cout<<endl;
cout<<"0:0"<<endl;
}
for(it;it!=str.end();it++)
{
if((xiaohua>=11||enemy>=11) && fabs(xiaohua-enemy)>=2 )
{
cout<<xiaohua<<":"<<enemy<<endl;
xiaohua=0;
enemy=0;
}
if(*it=='W') xiaohua++;
if(*it=='L') enemy++;
if((it==(str.end()-1)) && (xiaohua>=11||enemy>=11)&& fabs(xiaohua-enemy)>=2)
{
cout<<xiaohua<<":"<<enemy<<endl;
cout<<"0:0"<<endl;
continue;
}
if(it==(str.end()-1))
{
cout<<xiaohua<<":"<<enemy<<endl;
}
}
cout<<endl;
it=str.begin();
xiaohua=0;
enemy=0;
for(it;it!=str.end();it++)
{
if((xiaohua>=21||enemy>=21) && fabs(xiaohua-enemy)>=2 )
{
cout<<xiaohua<<":"<<enemy<<endl;
xiaohua=0;
enemy=0;
}
if(*it=='W') xiaohua++;
if(*it=='L') enemy++;
if((it==(str.end()-1)) && (xiaohua>=21||enemy>=21)&& fabs(xiaohua-enemy)>=2)
{
cout<<xiaohua<<":"<<enemy<<endl;
cout<<"0:0"<<endl;
continue;
}
if(it==(str.end()-1))
{
cout<<xiaohua<<":"<<enemy<<endl;
}
}
return 0;
}
这道题并没有给出数据范围，用string可以不用考虑数组开多大。emmmmm，尤其要注意特殊情况，嗯嗯，就是0：0的问题。我的特殊情况处理可能有点差劲，去看看其他人的代码 -
0
#include <iostream> #include <cmath> using namespace std; int main() { char a[80000]={'0'}; char tmp; int count=0; cin >> tmp; while(!(tmp=='E')) if (!(tmp=='\n') && !(tmp=='\0')) { a[count]=tmp; count++; cin >> tmp; } int ptr, cw, cl, k, t=0; k=11; LABEL: ptr=0; cw=0; cl=0; while (ptr!=count+1) { if (a[ptr]=='W') cw++; else if (a[ptr]=='L') cl++; ptr++; if ((cw>=k || cl>=k) && abs(cw-cl)>=2) { cout << cw << ':' << cl << endl; cw=0; cl=0; } } cout << cw << ':' << cl << endl << endl; t++; k=21; if(t==1) goto LABEL; return 0; } -
0
1、数组空间尽量大
2、比赛正好结束，要开始下一场比赛。#include <stdio.h> #include <math.h> #include <stdlib.h> void getScore(char *beg,int totle){ int red = 0, blue = 0; while(*beg!=69){ if(red>=totle||blue>=totle){ if(abs(red-blue)>=2){ printf("%d:%d\n",red,blue); red=0,blue=0; } } if(*beg==87) red++; else if(*beg==76) blue++; beg++; } printf("%d:%d\n",red,blue); if((red>=totle||blue>=totle)&&abs(red-blue)>=2) printf("0:0\n"); } int main() { char *letBeg; char *letAry = malloc(10000); //开辟大的空间正确，否则存在错误 scanf("%c",letAry); letBeg = letAry; while(*letAry!=69){ if(*letAry==87||*letAry==76){ letAry++; } scanf("%c",letAry); } getScore(letBeg,11); printf("\n"); getScore(letBeg,21); return 0; } -
0
1、数组空间尽量的大，100000
2、注意开始直接输入E的情况
3、最后一个报错，是因为没有考虑正好结束，要开始下一场比赛。#include <stdio.h> #include <math.h> #include <stdlib.h> void getScore(char *beg,int totle){ int red = 0, blue = 0; while(*beg!=69){ if(red>=totle||blue>=totle){ if(abs(red-blue)>=2){ printf("%d:%d\n",red,blue); red=0,blue=0; } } if(*beg==87) red++; else if(*beg==76) blue++; beg++; } printf("%d:%d\n",red,blue); if((red>=totle||blue>=totle)&&abs(red-blue)>=2) printf("0:0\n"); } int main() { char *letBeg; char *letAry = malloc(10000); //开辟大的空间正确，否则存在错误 scanf("%c",letAry); letBeg = letAry; while(*letAry!=69){ if(*letAry==87||*letAry==76){ letAry++; } scanf("%c",letAry); } getScore(letBeg,11); printf("\n"); getScore(letBeg,21); return 0; } -
0
非递归的写法,刚开始res数组500不能AC,改成了10000才行
思路是设置两个数组res11,res22分别储存比分,
两个指针 p11 , p22代表当前场次编号
在读到'E'之前,每次循环读入一个字符,并进行判断,若比赛结束指针+1// // Created by eh5 on 17-11-7. // #include <stdio.h> #include <stdlib.h> struct sc { int w; int l; }; int state(int a,int b,int sc){ if(a<sc && b<sc) return 0; else if(abs(a-b) >1) return 1; return 0; } int main(){ int C; struct sc res11[10000],res21[10000]; int p11=0,p21=0; while((C=getchar())!='E'){ if((C != 'W') && (C != 'L')) continue; if(C == 'W'){ res11[p11].w += 1; res21[p21].w += 1; } else { res11[p11].l += 1; res21[p21].l += 1; } if(state(res11[p11].w ,res11[p11].l,11)){ p11++; } if(state(res21[p21].w ,res21[p21].l,21)){ p21++; } } for(int i=0;i<=p11;i++){ printf("%d:%d\n",res11[i].w,res11[i].l); } printf("\n"); for(int i=0;i<=p21;i++){ printf("%d:%d\n",res21[i].w,res21[i].l); } } -
0
electrodynamix LV 9 @ 2017-11-07 18:54:36
小学数学题
没什么可说的
代码：
dalao不喜勿喷#include<cstdio> #include<cstring> #include<algorithm> int L_11[20001],W_11[20001],L_21[20001],W_21[20001],len_11=1,len_21=1; int main(){ char ch; ch=getchar(); while(ch!='E') { if(ch=='W') W_11[len_11]++; else if(ch=='L') L_11[len_11]++; if((W_11[len_11]>=11 && W_11[len_11]-L_11[len_11]>=2) || (L_11[len_11]>=11 && L_11[len_11]-W_11[len_11]>=2)) len_11++; if(ch=='W') W_21[len_21]++; else if(ch=='L') L_21[len_21]++; if((W_21[len_21]>=21 && W_21[len_21]-L_21[len_21]>=2) || (L_21[len_21]>=21 && L_21[len_21]-W_21[len_21]>=2)) len_21++; ch=getchar(); } for(int i=1;i<=len_11;i++) printf("%d:%d\n",W_11[i],L_11[i]); printf("\n"); for(int i=1;i<=len_21;i++) printf("%d:%d\n",W_21[i],L_21[i]); return 0; } -
0
难点。。。一在规则，二在读入
给新手的建议：
1.能用cin别用其他的，除非数据特别大，你永远不知道getchar之类的会造成什么后果（洛谷是卡getchar的，亲测）
2.大部分在线OJ，对于有多个输出的题目，可以即时输出，但是为了保险，一般存储完最后输出（一些交互式题目或云端评测除外）
3.好好读题，信竞从来就不是纯理科。。。到后来就会体会到读不懂题的恐惧
贴两组代码，一组是刚入坑的代码，一组是最近写的，没想到思路居然还惊人相似呢。。。
NOIP备战最后几天，考完之后不知道还会不会学，看将来吧。。。
刚入坑时代码#include<iostream> #include<cstdio> #include<cstring> using namespace std; int l[2][10000];int m[2][100000]; int main() { char a='c'; int x=0,y=0,u=0,v=0,i=0,k=0; for(;a!='E';) { a=getchar(); if(a=='W') { x++;u++; } if(a=='L') { y++;v++; } if((x>=11&&x-y>=2)||(y>=11&&y-x>=2)) { m[0][k]=x;m[1][k]=y;k++;x=0;y=0; } if((u>=21&&u-v>=2)||(v>=21&&v-u>=2)) { l[0][i]=u;l[1][i]=v;i++;u=0;v=0; } } for(int p=0;p<k;p++) cout<<m[0][p]<<":"<<m[1][p]<<endl; cout<<x<<":"<<y<<endl; cout<<endl; for(int h=0;h<i;h++) cout<<l[0][h]<<":"<<l[1][h]<<endl; cout<<u<<":"<<v; return 0; }现在代码
#include<iostream> #include<cstdio> #include<cmath> using namespace std; int ans[300][2],ans2[300][2]; int main() { char c; int sum1=0,sum2=0,sum3=0,sum4=0,zong=0,gong1=0,gong2=0,i; while(cin>>c) { if(c=='E') { gong1++;gong2++; ans[gong1][0]=sum1; ans[gong1][1]=sum2; ans2[gong2][0]=sum3; ans2[gong2][1]=sum4; break; } if(c=='W') { sum1++; sum3++; } if(c=='L') { sum2++; sum4++; } if((sum1>=11||sum2>=11)&&abs(sum1-sum2)>=2) { gong1++; ans[gong1][0]=sum1; ans[gong1][1]=sum2; sum1=0; sum2=0; } if((sum3>=21||sum4>=21)&&abs(sum3-sum4)>=2) { gong2++; ans2[gong2][0]=sum3; ans2[gong2][1]=sum4; sum3=0; sum4=0; } } for(i=1;i<=gong1;i++) cout<<ans[i][0]<<":"<<ans[i][1]<<endl; cout<<endl; for(i=1;i<=gong2;i++) cout<<ans2[i][0]<<":"<<ans2[i][1]<<endl; return 0; } -
0
我这个是用java写的，看了一下别的题解都没有java的 所以我来上传一下。用String和StringBuilder都可以，字符串简单点。代码如下：
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
StringBuilder s=new StringBuilder();
// String s =new String();
int x=0; //第一次出现E的位置
while(true)//while(s.charAt(x)!='E')
{
s.append(sc.nextLine());
// s+=sc.nextLine();
x=s.indexOf("E");
if(x!=-1)break;
}
int hua=0;
int huad=0;
int hua2=0;
int huad2=0;for (int i = 0; i<x; i++) {
if(s.charAt(i)=='W')
hua++;
if(s.charAt(i)=='L')
huad++;
if((hua==11&&huad==0)||(hua==0&&huad==11)||(hua>=11&&(hua-huad)>=2)
|| (huad>=11&&(huad-hua)>=2))
{
System.out.println(hua+":"+huad);
hua=0;
huad=0;
}
}
System.out.println(hua+":"+huad);
System.out.println();for (int i = 0; i < x; i++) {
if(s.charAt(i)=='W')
hua2++;
if(s.charAt(i)=='L')
huad2++;
if((hua2==21&&huad2==0)||(hua2==0&&huad2==21)||(hua2>=21&&(hua2-huad2)>=2)
|| (huad2>=21&&(huad2-hua2)>=2))
{
System.out.println(hua2+":"+huad2);
hua2=0;
huad2=0;
}
}
System.out.println(hua2+":"+huad2);
}
} -
0
2351599120 LV 9 @ 2017-10-24 06:34:09
来个不用字符串的
#include<bits/stdc++.h> using namespace std; int aa[10001],bb[10001]; int main() { char a; int l=0,r=0,ll=0,rr=0,s=1; while(1) { scanf("%c",&a); if(a=='E') { printf("%d:%d\n",l,r); aa[s]=ll; bb[s]=rr; printf("\n"); for(int i=1;i<=s;i++) { printf("%d:%d\n",aa[i],bb[i]); } return 0; } if(a=='W') { l++; ll++; } if(a=='L') { r++; rr++; } if((l>=11||r>=11)&&(l-r>=2||r-l>=2)) { printf("%d:%d\n",l,r); l=0; r=0; } if((ll>=21||rr>=21)&&(ll-rr>=2||rr-ll>=2)) { aa[s]=ll; bb[s]=rr; s++; ll=0; rr=0; } } } -
0
新手python，没动脑子。
sba=0
sbb=0
nba=0
nbb=0
shuju=[]
ansa=[]
ansb=[]
while True :
s=raw_input()
if "E" not in s:
shuju.append(s)
else:
shuju.append(s[:s.find("E")])
break
for a in shuju:
for i in range(len(a)):
if a[i]=="W":
sba+=1
nba+=1
elif a[i]=="L":
sbb+=1
nbb+=1
if (sba>=11 or sbb>=11) and abs(sba-sbb)>=2:
ansa.append(str(sba)+":"+str(sbb))
sba=0
sbb=0
if (nba>=21 or nbb>=21) and abs(nba-nbb)>=2:
ansb.append(str(nba)+":"+str(nbb))
nba=0
nbb=0
ansa.append(str(sba)+":"+str(sbb))
ansb.append(str(nba)+":"+str(nbb))
for i in ansa:
print i
print ""
for i in ansb:
print i
蛟川洋
LV 5
@ 2018-03-25 18:41:13
