[pj2015题解]其实是纯代码.

 1 #include <iostream>
 2 #include <fstream>
 3 #include <cstdlib>
 4 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
 5 using namespace std;
 6 
 7 ifstream fin("coin.in");
 8 ofstream fout("coin.out");
 9 
10 int cnt_shu;
11 long long he=0;
12 
13 int main(int argc, char** argv) {
14 fin>>cnt_shu;
15 int shi=0;
16 for(int x=1;x<=cnt_shu;x+=shi){
17  shi++;
18  if(x+shi-1<=cnt_shu)he+=shi*shi;
19  else he+=(cnt_shu-x+1)*shi;  
20 }
21 fout<<he;
22 return 0;
23 }

 1 #include <iostream>
 2 #include <fstream>
 3 #include <cstdlib>
 4 #include <string>
 5 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
 6 using namespace std;
 7 
 8 ifstream fin("mine.in");
 9 ofstream fout("mine.out");
10 
11 int cnt_hang=0,cnt_lie=0;
12 int jv[105][105];
13 int hez[8]={1,-1,0,0,1,-1,1,-1};
14 int zoz[8]={0,0,1,-1,1,-1,-1,1};    
15 
16 int zhao(int he,int zo){
17 
18 int ans=0;
19 for(int x=0;x<8;x++){
20  int han=he+hez[x];
21  int zon=zo+zoz[x];
22  if(he<1||zo<1||he>cnt_hang||zo>cnt_lie)continue;
23  if(jv[han][zon]==0)continue;
24  else ans++;
25 }
26 return ans;
27 }
28 
29 
30 int main(int argc, char** argv) {
31 fin>>cnt_hang>>cnt_lie;
32 for(int x=1;x<=cnt_hang;x++){
33  string a;fin>>a;
34  for(int y=1;y<=cnt_lie;y++)if(a[y-1]==‘*‘)jv[x][y]=1;    
35 }
36 for(int x=1;x<=cnt_hang;x++){
37  for(int y=1;y<=cnt_lie;y++){
38   if(jv[x][y]==0)fout<<zhao(x,y);    
39   else fout<<"*";
40  }    
41  fout<<endl;
42 }
43 return 0;
44 }

 1 #include <iostream>
 2 #include <fstream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
 6 using namespace std;
 7 
 8 ifstream fin("sum.in");
 9 ofstream fout("sum.out");
10 
11 int cnt_ge,cnt_col=0;
12 int id_num[100005];
13 int head[100005],cnt=0;
14 int zhan[3][100005];
15 int zhan2[100005];
16 long long ans;
17 struct lian{
18  int nxt;
19  int to;    
20 };
21 lian cun[100005];
22 
23 void add(int sze,int yan);
24 void sou(int yan);
25 void suan_he(int ces);
26 
27 
28 void add(int sze,int yan){
29 cnt++;
30 cun[cnt].to=sze;
31 cun[cnt].nxt=head[yan];
32 head[yan]=cnt;
33 return;
34 }
35 
36 
37 int gs1=0,gs2=0;
38 long long he1=0ll,he2=0ll;
39 void sou(int yan){
40 memset(zhan,0,sizeof(zhan));
41 int sze=cun[head[yan]].to;
42 gs1=0,gs2=0;
43 he1=0ll,he2=0ll;
44 for(int x=head[yan];x!=-1;x=cun[x].nxt){    
45  int dao=cun[x].to;
46  if((dao+sze)%2==0){
47   he1+=dao;gs1++;
48   zhan[0][gs1]=dao;    
49  }
50  else{
51   he2+=dao;gs2++;
52   zhan[1][gs2]=dao;    
53  }     
54 }
55 
56 for(int x=0;x<2;x++)suan_he(x);    
57 return;
58 }
59 
60 
61 void suan_he(int ces){
62 int gs=0;long long he=0ll;
63 if(ces==0){gs=gs1;he=he1;}     
64 else {gs=gs2;he=he2;}
65 if(gs==0||gs==1)return;
66 int shu=0;
67 for(int x=gs;x>=1;x--){
68  shu=((he+zhan[ces][x])/2*id_num[zhan[ces][x]])*10007;
69  ans+=shu*2%10007;
70  ans*=10007;
71 }
72 
73 return;
74 }
75 
76 
77 int main(int argc, char** argv) {
78 fin>>cnt_ge>>cnt_col;
79 memset(head,-1,sizeof(head));
80 for(int x=1;x<=cnt_ge;x++)fin>>id_num[x];    
81 for(int x=1;x<=cnt_ge;x++){
82 int yan;fin>>yan;
83 add(x,yan);    
84 }
85 for(int x=1;x<=cnt_col;x++){
86  sou(x);    
87 }
88 fout<<ans%10007;
89 return 0;
90 }

 1 #include <iostream>
 2 #include <fstream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 /* run this program using the console pauser or add your own getch, system("pause") or input loop */
 6 using namespace std;
 7 
 8 ifstream fin("salesman.in");
 9 ofstream fout("salesman.out");
10 
11 int cnt_shu;
12 int jv[100005];
13 int a[100005];
14 int ans[100005];
15 int shang[100005];
16 int hou[100005];
17 int gs=0;
18 int dui[100005];
19 int wei;
20 void pout(int sze);
21 void zhuan( );
22 void na(int sze);
23 
24 void pout(int sze){
25  gs++;dui[gs]=sze;
26  int now=gs,fu;    
27  while(now/2>0){
28  fu=now/2; 
29  if(a[dui[fu]]<a[dui[now]])return;    
30  int b=dui[fu];dui[fu]=dui[now];dui[now]=b;
31  now=fu;
32  }
33  return;
34 }
35 
36 void zhuan( ){
37 for(int x=2;x<=cnt_shu;x++)shang[x]=x-1;
38 for(int x=1;x<cnt_shu;x++)hou[x]=x+1;
39 for(int x=1;x<cnt_shu;x++){
40  int dai1=a[dui[1]];
41  int dai2=(jv[wei]-jv[shang[wei]])*2+a[wei];
42  if(dai1>=dai2){
43   ans[cnt_shu-x]=ans[cnt_shu-x+1]-dai2;
44   wei=shang[wei]; na(wei);
45  }     
46  else{
47   hou[shang[dui[1]]]=hou[dui[1]];
48   shang[hou[dui[1]]]=shang[dui[1]];
49   na(1);
50   ans[cnt_shu-x]=ans[cnt_shu-x+1]-dai1;    
51    
52 }
53  }
54 return;    
55 }
56 
57 
58 void na(int sze){
59 dui[sze]=dui[gs];
60 gs--;    
61 int now=sze,zi=0;
62 while(now*2<=gs){
63  zi=now*2;
64  if(a[dui[zi+1]]<a[dui[zi]]&&now*2<gs)zi++;
65  if(a[dui[zi]]>a[dui[now]])return;
66  int b=dui[zi];dui[zi]=dui[now];
67  dui[now]=b;now=zi;
68 }
69 return;
70 }
71 
72 
73 int main(int argc, char** argv) {
74 fin>>cnt_shu;
75 for(int x=1;x<=cnt_shu;x++){
76 fin>>jv[x];ans[cnt_shu]=jv[x]*2;    
77 }
78 for(int x=1;x<=cnt_shu;x++){
79 fin>>a[x];
80 if(x!=cnt_shu)pout(x);
81 ans[cnt_shu]+=a[x];
82 }
83 wei=cnt_shu;
84 zhuan( );
85 for(int x=1;x<=cnt_shu;x++)fout<<ans[x]<<endl;
86 
87  return 0;
88 }