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這個(gè)題目A的很傻，一次接入三個(gè)字符，然后自己不停的來處理各種情況的效果，呵呵，各種格式是應(yīng)該注意的，空格算是小陷阱？呵呵，所以一般復(fù)制那個(gè)sample再改就好了，freopen("a.in","r",stdin);freopen("a.out","w",stdout);這種文件讀入讀出的方式對(duì)觀察自己的輸出結(jié)果很有好處，推薦使用防止PE，呵呵，也許是北大的數(shù)據(jù)比較弱，這個(gè)同樣的題目在天大的OJ就過不了~WA的很郁悶。。。
1 Source Code
2
3Problem: 3337 User: hongtaozhy
4Memory: 296K Time: 0MS
5Language: G++ Result: Accepted
6
7Source Code
8#include<stdio.h>
9#include<string.h>
10char res[10000];
11char fes[10000];
12  char zd[26]= {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};
13bool mark[26];
14int main(){
15 int n;
16 int sum;
17 int key ;
18 //freopen("g.in","r",stdin);
19 //freopen("gg.in","w",stdout);
20 scanf("%d\n",&n);
21 int len ;
22  while(n--){
23 key = 0 ;
24 memset(mark,0,sizeof(mark));
25 for(int i = 0 ;i < 26 ; i++ )
26 zd[i]=i+1;
27
28 sum = 0 ;
29 gets(fes);
30 printf("Expression: %s\n",fes);
31
32
33 len = strlen ( fes );
34 int t = 0 ;
35 for(int i = 0 ;i < len ; i++ ){
36 if(fes[i]!=' ') res[t++]=fes[i];
37 if(fes[i]=='\0') { res[t++]=0;res[t++]=0;}
38 }
39 // printf("%s\n",res);
40 len = strlen ( res );
41 for( int i = 0 ; i < len ; i++ ){
42 if(res[i]<='z'&&res[i]>='a'&&res[i+1]==res[i+2]){
43 if(key == 0)
44 sum += zd[res[i]-'a'];
45 else
46 sum -= zd[res[i]-'a'];
47 if(res[i+1]=='-')
48 zd[res[i]-'a']--;
49 else if(res[i+1]=='+')
50 zd[res[i]-'a']++;
51 mark[res[i]-'a']=1;
52
53 i+=2;
54 continue;
55 }
56 else if(res[i+2]<='z'&&res[i+2]>='a'&&res[i+1]==res[i]){
57 if(res[i+1]=='-')
58 zd[res[i+2]-'a']--;
59 else if(res[i+1]=='+')
60 zd[res[i+2]-'a']++;
61 if(key == 0)
62 sum += zd[res[i+2]-'a'];
63 else
64 sum -= zd[res[i+2]-'a'];
65 mark[res[i+2]-'a']=1;
66
67 i+=2;
68 continue;
69 }
70 else if(res[i]<='z'&&res[i]>='a'){
71
72 if(key == 0)
73 sum += zd[res[i]-'a'];
74 else
75 sum -= zd[res[i]-'a'];
76 mark[res[i]-'a']=1;
77 if(res[i+1]=='+') key=0;
78 else key=1;
79 }
80 else if(res[i]=='-'&&res[i+1]=='-'&&res[i+2]=='+')
81 key=0;
82 else if(res[i]=='+'&&res[i+1]=='+'&&res[i+2]=='-')
83 key=1;
84 else if(res[i]=='-'&&res[i+1]=='+'&&res[i+2]=='+')
85 key=1;
86 else if(res[i]=='+'&&res[i+1]=='-'&&res[i+2]=='-')
87 key=0;
88 else if(res[i]=='+'&&res[i+1]<='z'&&res[i+1]>='a')
89 key=0;
90 else if(res[i]=='-'&&res[i+1]<='z'&&res[i+1]>='a')
91 key=1;
92 else if(res[i]=='+'&&res[i+1]=='+'&&res[i+2]=='+')
93 key=0;
94 else if(res[i]=='-'&&res[i+1]=='-'&&res[i+2]=='-')
95 key=1;
96 }
97
98
99 printf("value = %d\n",sum);
100 for(int i = 0 ; i < 26 ; i ++ ){
101 if(mark[i]!=0)
102 printf("%c = %d\n",i+'a',zd[i]);
103 }
104 }
105 // while(1);
106return 0 ;
107 }
108
109
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