A：因为是三位太太共同的 所以要平摊。
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 #include <iostream>
6 #include <cmath>
7 #include <queue>
8 #include <map>
9 #include <stack>
10 #include <list>
11 #include <vector>
12 #include <ctime>
13 #define LL __int64
14 #define eps 1e-8
15 using namespace std;
16 int jin(double x)
17 {
18 if(x - x/1 < 0.5)
19 return x/1 ;
20 else return x/1 +1;
21 }
22 int main()
23 {
24
25 int n ;
26 scanf("%d",&n);
27 for(int i = 1;i <= n;i++)
28 {
29 int x, y ,z ;
30 scanf("%d %d %d",&x,&y,&z);
31
32
33 x = jin(1.0*x * z/(x+y) + 1.0*(x-y) *z/(x+y));
34 y = z - x;
35 printf("%d\n",x);
36 }
37 return 0;
38 }
B：水模拟
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 #include <iostream>
6 #include <cmath>
7 #include <queue>
8 #include <map>
9 #include <stack>
10 #include <list>
11 #include <vector>
12 #include <ctime>
13 #define LL __int64
14 #define eps 1e-8
15 using namespace std;
16 char str[10004];
17 int sta[100005];
18 int change(int l,int r)
19 {
20 int ans = 0 ;
21 for(int i= l; i <= r ; i ++)
22 {
23 ans *= 10 ;
24 ans += str[i] - ‘0‘;
25 }
26 return ans;
27 }
28 int main()
29 {
30 int t ;
31 scanf("%d",&t);
32 while(t--)
33 {
34 int n ;
35 int s = 0 ;
36 scanf("%d",&n);
37 for(int i = 1;i <= n;i++)
38 {
39 scanf("%s",str);
40 if(str[0] == ‘L‘)
41 {
42 sta[i] = 1;
43 s -- ;
44 }
45 else if(str[0] == ‘R‘)
46 {
47 sta[i] = 0 ;
48 s ++;
49 }
50 else {
51
52 //int k = change(8,p);
53 int k ;
54 scanf("%s",str);
55 scanf("%d",&k);
56 //printf("%d\n",k);
57 sta[i] = sta[k];
58 if(sta[i])
59 {
60 s -- ;
61 }else s ++ ;
62 }
63 }
64 printf("%d\n",s);
65 }
66 return 0 ;
67 }
C：细节题 ，用map代码量可能会少一点
1 // File Name: c.cpp
2 // Author: darkdream
3 // Created Time: 2014年10月07日 星期二 10时06分00秒
4
5 #include<vector>
6 #include<list>
7 #include<map>
8 #include<set>
9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25
26 using namespace std;
27 char bstr[1000];
28 char str[200][5][200];
29 char str1[200][5][200];
30 int lstr;
31 int lstr1;
32 void solve(char s[][5][200],int &num)
33 {
34 int p = 0 ;
35 if(bstr[1] == ‘}‘)
36 return;
37
38 while(bstr[p] != ‘}‘)
39 {
40 //printf("%d\n",p);
41 num ++ ;
42 int i ;
43 p ++ ;
44 for(i = p ;bstr[i] != ‘:‘ ;i ++)
45 {
46 s[num][1][i-p] = bstr[i];
47 }
48 //printf("%d %s",i,s[num][1]);
49 s[num][1][i-p] = ‘\0‘;
50 p = i+1 ;
51 //printf(" %s\n",s[num][1]);
52
53 for(i = p ;bstr[i] != ‘,‘&& bstr[i] != ‘}‘;i ++)
54 {
55 s[num][2][i-p] = bstr[i];
56 }
57 // printf("%d\n",p);
58 s[num][2][i-p] = ‘\0‘;
59 p = i;
60 }
61 }
62 int srt[200];
63 int srt1[200];
64 int cmp(int a,int b)
65 {
66 if(strcmp(str[a][1],str[b][1]) < 0)
67 return 1;
68 return 0 ;
69 }
70 int cmp1(int a,int b)
71 {
72 if(strcmp(str1[a][1],str1[b][1]) < 0)
73 return 1;
74 return 0 ;
75 }
76 char *push[200];
77 char *change[200];
78 char *sub[200];
79 int lp,lc,ls;
80 int solve()
81 {
82 lp = lc = ls = 0;
83 int p = 1 ;
84 for(int i = 1; i <= lstr1 ;i ++)
85 {
86 int ok = 0 ;
87 int x = srt1[i];
88 for(int j = p;j <= lstr; j ++)
89 {
90 int y = srt[j];
91 if(strcmp(str[y][1],str1[x][1]) == 0 )
92 {
93 for(int s = p ;s < j;s ++)
94 {
95 ls ++;
96 sub[ls] = str[srt[s]][1];
97 //printf("%s\n",str[srt[p]][1]);
98 }
99 p = j+1;
100 if(strcmp(str[y][2],str1[x][2]) != 0 )
101 {
102 lc ++;
103 change[lc] = str[y][1];
104 }
105 ok = 1;
106 break;
107 }
108 }
109 if(!ok)
110 {
111 lp ++ ;
112 push[lp] = str1[x][1];
113 //printf(" %d %s************************\n",x,str1[x][1]);
114 }
115 }
116 for(;p <= lstr; p ++)
117 {
118 ls ++ ;
119 sub[ls] = str[srt[p]][1];
120 //printf("%s\n",str[srt[p]][1]);
121 }
122 if(lp != 0 )
123 {
124 printf("+");
125 for(int i = 1;i <= lp ;i ++)
126 printf(i == 1?"%s":",%s",push[i]);
127 printf("\n");
128 }
129 if(ls != 0 )
130 {
131 printf("-");
132 for(int i = 1;i <= ls ;i ++)
133 printf(i == 1?"%s":",%s",sub[i]);
134 printf("\n");
135 }
136 if(lc != 0 )
137 {
138 printf("*");
139 for(int i = 1;i <= lc ;i ++)
140 printf(i == 1?"%s":",%s",change[i]);
141 printf("\n");
142 }
143 if(lp == 0 && ls == 0 && lc == 0 )
144 {
145 return 0;
146 }
147 return 1;
148 }
149 int main(){
150 int n ;
151 scanf("%d",&n);
152 while(n--)
153 {
154 lstr = 0 ;
155 lstr1 = 0 ;
156 scanf("%s",bstr);
157 solve(str,lstr);
158 /*for(int i = 1;i <= lstr;i ++)
159 {
160 printf("%s %s ***",str[i][1],str[i][2]);
161 }
162 printf("***\n");*/
163 scanf("%s",bstr);
164 solve(str1,lstr1);
165 /*for(int i = 1;i <= lstr1;i ++)
166 {
167 printf("%s %s ***",str1[i][1],str1[i][2]);
168 }*/
169 for(int i =1;i <= lstr;i ++)
170 srt[i] = i ;
171 sort(srt+1,srt+1+lstr,cmp);
172 for(int i =1;i <= lstr1;i ++)
173 srt1[i] = i ;
174 sort(srt1+1,srt1+1+lstr1,cmp1);
175 if(!solve())
176 printf("No changes\n");
177 printf("\n");
178 }
179 return 0;
180 }
D：从小数点后枚举这一位 是 0 ，还是 1 装换成大数浮点形 平方以后和 n 比较，确定这一位 ，一直确定到 130位就行
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <sstream>
5 #include <algorithm>
6 #include <cmath>
7 using namespace std;
8 int a1[10010],a2[10010],b[10000],c[10000],l1,l2,s[110000];
9 int l3,l4,l5,point,point2,point3,point1;
10 char ans[1000];
11 char ss[23];
12 int chengfa()
13 {
14 int pos,i,j;
15 memset(s,0,sizeof(s));
16 for (i=1;i<=l4;i++)
17 for (j=1,pos=i;j<=l4;j++)
18 s[pos++]+=a2[i]*a2[j];
19 pos-=1;
20 for (i=1;i<=pos;i++)
21 if (s[i]>=10)
22 {
23 if (i==pos) pos++;
24 s[i+1]+=s[i]/10;
25 s[i]%=10;
26 }
27 return pos;
28 }
29 void jia()
30 {
31 int p=1,i;
32 if (point > point1)
33 {
34 for (i=1;i<=point - point1;i++)
35 a2[p++]=a1[i];
36 int tt=1;
37 for (i=point - point1+1;i<=l1;i++)
38 a2[p++]=a1[i]+c[tt++];
39 point2=point;
40 }
41 else
42 {
43 for (i=1;i<=point1-point;i++)
44 a2[p++]=c[i];
45 int tt=i;
46 for (i=1;i<=l1;i++)
47 a2[p++]=a1[i]+c[tt++];
48 point2=point1;
49 }
50 int kk=0;
51 p--;
52 for (i=1;i<=p;i++)
53 {
54 a2[i]+=kk;
55 kk=a2[i]/10;
56 a2[i]%=10;
57 }
58 if (kk!=0) a2[++p]=kk;
59 l4=p;
60 }
61 int gobj()
62 {
63 int sl=l5,bl=l2;
64 if (sl-point3>bl) return 1;
65 else if (sl-point3<bl) return -1;
66 while (sl>0 && bl>0)
67 {
68 if (s[sl]>b[bl]) return 1;
69 if (s[sl]<b[bl]) return -1;
70 sl--;bl--;
71 }
72 if (sl==0) return 0;
73 else return 1;
74 }
75 int main()
76 {
77 int T,n;
78 scanf("%d",&T);
79 while (T--)
80 {
81 memset(ans,0,sizeof(ans));
82 memset(a1,0,sizeof(a1));
83 memset(a2,0,sizeof(a2));
84 memset(c,0,sizeof(c));
85 memset(b,0,sizeof(b));
86 scanf("%d",&n);
87 getchar();
88 scanf("%s",&ss);
89 int m=sqrt(n);
90 int mm=m;
91 int j=1;
92 l1=0;
93 point=0;
94 while (mm)
95 {
96 a1[j++]=mm % 10;
97 mm/=10;
98 l1++;
99 }
100 point=0;
101 mm=n;
102 j=1;l2=0;
103 while (mm)
104 {
105 b[j++]=mm%10;
106 mm/=10;
107 l2++;
108 }
109 c[1]=1;l3=1;
110 int i;
111 for (i=1;i<=130;i++)
112 {
113 for (j=1;j<=l3;j++)
114 c[j]*=5;
115 int kk=0;
116 for (j=1;j<=l3;j++)
117 {
118 c[j]+=kk;
119 kk=c[j]/10;
120 c[j]=c[j]%10;
121 }
122 if (kk) c[++l3]=kk;
123 point1=i;
124 jia();
125 /*for (j=l4;j>0;j--)
126 {
127 if (point2==j) cout<<".";
128 printf("%d",a2[j]);
129 }
130 cout<<endl;*/
131 l5=chengfa(); //平方后长度
132 point3=2*point2; //平方后小数点位置
133 /*for (j=l5;j>0;j--)
134 {
135 if (point3==j) cout<<".";
136 printf("%d",s[j]);
137 }
138 cout<<endl;*/
139 int re=gobj();
140 if (re==1) //1:s>b -1:s<b
141 ans[i-1]=‘0‘;
142 else if (re==-1)
143 {
144 ans[i-1]=‘1‘;
145 memset(a1,0,sizeof(a1));
146 for (j=1;j<=l4;j++)
147 a1[j]=a2[j];
148 l1=l4;
149 point=point2;
150 }
151 else break;
152 }
153 for (j=i;j<=130;j++)
154 ans[i]=‘0‘;
155 ans[130]=‘\0‘;
156 //cout<<ans<<endl;
157 cout<<strstr(ans,ss) - ans << endl;
158 }
159 return 0;
160 }
E：这题字典树或者 字符串排序（只对下标排序都行 见：http://www.cnblogs.com/zyue/p/4007476.html）
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 #include <iostream>
6 #include <cmath>
7 #include <queue>
8 #include <map>
9 #include <stack>
10 #include <list>
11 #include <vector>
12 #include <ctime>
13 #define LL __int64
14 #define eps 1e-8
15 using namespace std;
16 string str[1005];
17 int b[1005];
18 bool cmp(int a, int b)
19 {
20 return str[a] < str[b];
21 }
22 int main()
23 {
24 int t ;
25 scanf("%d",&t);
26 while(t--)
27 {
28 int n ;
29 scanf("%d",&n);
30 for(int i = 1;i <= n;i ++)
31 cin >> str[i] ;
32 for(int i =1;i <= n;i ++)
33 b[i] = i ;
34 sort(b+1,b+1+n,cmp);
35 /* for(int i = 1;i <= n;i ++)
36 printf("%d ",b[i]);
37 printf("\n");
38 */
39 int k = 0 ;
40 int tk ;
41 int ans = 0 ;
42 for(int i = 2;i <= n;i ++)
43 {
44 int len1 = str[b[i-1]].size();
45 int len2 = str[b[i]].size();
46 for(int j = 0 ;j < len1;j++)
47 {
48 if(str[b[i-1]][j] != str[b[i]][j])
49 {
50 tk = j + 1;
51 break;
52 }
53 }
54 ans += max(k,tk);
55 // printf("%d %d %d\n",k,tk,ans);
56 k = tk;
57
58 }
59 ans += k ;
60 printf("%d\n",ans);
61 }
62 return 0 ;
63 }
F：这个题我们可以看到 点数只有 16个点 我们可以 二进制枚举出要选点所有的情况以后 然后再用prime求出一种情况的答案。比较求得最大值
不过这里有个地方要注意，两点之间可能有多条边，如果是链接矩阵的话要判断一下再加边入阵。
1 #include<cstdio>
2 #include<cstring>
3 #include<map>
4 #include<vector>
5 #include<cmath>
6 #include<cstdlib>
7 #include<stack>
8 #include<queue>
9 #include <iomanip>
10 #include<iostream>
11 #include<algorithm>
12 using namespace std ;
13 const int inf = 1<<30 ;
14
15 int n,m,k,tmp;
16 int a[100],g[50][50],dist[100],vis[100],s[100];
17
18
19 int work1()
20 {
21 int now ,pos=0,cost=0,MIN=inf,num;
22 //memset(vis,0,sizeof(vis));
23 for(int i = 0 ; i < tmp; i++ )
24 {
25 now =s[i] ;
26 dist[now] = g[1][now] ;
27 vis[now]=0;
28 // printf("%d ",dist[now]) ;
29 } // puts("");
30 vis[1]=1;num=0;
31 for(int i = 1 ; i < tmp ; i++)
32 {
33 MIN=inf ;
34 for(int j = 0 ; j < tmp;j++)
35 {
36 now = s[j] ;
37 if(MIN > dist[now] && !vis[now])
38 {
39 MIN=dist[now] ;
40 pos=now ;
41 }
42 }
43 cost += MIN ;
44 vis[pos]=1;
45 for(int j = 0 ; j < tmp; j++)
46 {
47 now= s[j] ;
48 if(dist[now] > g[pos][now] && !vis[now])
49 dist[now] = g[pos][now] ;
50 }
51 }
52 int sum = 0;
53 for(int i = 0 ; i < tmp ; i++)
54 { now = s[i] ;
55 sum += a[now] ;
56 if(vis[now]) num++;
57 }
58 if(cost <= k && num == tmp)
59 return sum ;
60 else return 0 ;
61 }
62
63 int print_subset(int x)
64 {
65 tmp =0 ;
66 for(int i=0;i<n;i++)
67 if(x&(1<<i))
68 {
69 s[tmp++]=i+1;
70
71 }
72 if(s[0]!=1 )
73 return 0;
74 else return 1;
75 }
76
77 int main()
78 {
79 int t ,u,v,w;
80 scanf("%d",&t) ;
81 while(t--)
82 {
83 scanf("%d%d%d",&n,&m,&k);
84 for(int i = 1 ; i <= n ; i++)
85 for(int j = 1 ; j <= n ;j++)
86 g[i][j]=inf ;
87 for(int i = 1 ; i <= n ; i++)
88 scanf("%d",&a[i]) ;
89
90 for(int i = 1 ; i <= m ; i++)
91 {
92 scanf("%d%d%d",&u,&v,&w) ;
93 if(g[u][v]>w)
94 g[u][v]=g[v][u]=w ;
95 }
96 int ans = 0 ;
97
98 for(int i=0;i<(1<<n);i++)
99 {
100 int flag = print_subset(i);
101 if(tmp==1)
102 {
103 ans = max(ans,a[1]) ;
104 continue;
105 }
106 int ans1 ;
107 if(flag )
108 {
109 ans1 = work1() ;
110 if(ans1 > ans)
111 {
112 ans = ans1;
113 /* for(int j = 0 ; j < tmp ; j++)
114 printf("%d ",s[j]) ;
115 puts(""); */
116 }
117
118 }
119 else continue ;
120
121 }
122
123 printf("%d\n",ans) ;
124 }
125 return 0;
126 }
J：注意到点只有 1000 ，所以直接暴力DP就行 dp[i] 表示为到a[i]这个点最长的公共递增长度。
每一次输入b[j]，从头开始枚举，a[i] < b[j] 且 长度最长的值，到了 b[j] = a[i],更新dp[i] 即可
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 #include <iostream>
6 #include <cmath>
7 #include <queue>
8 #include <map>
9 #include <stack>
10 #include <list>
11 #include <vector>
12 #include <ctime>
13 #define LL __int64
14 #define eps 1e-8
15 using namespace std;
16 int a[1004];
17 int b[1004];
18 int dp[1004];
19 int main()
20 {
21 int numa,numb;
22 int t;
23 scanf("%d",&t);
24 while(t--)
25 {
26 scanf("%d",&numa);
27 for(int i= 1;i<= numa;i ++)
28 scanf("%d",&a[i]);
29
30 scanf("%d",&numb);
31 memset(dp,0,sizeof(dp));
32 for(int i= 1;i<= numb;i ++)
33 {
34 scanf("%d",&b[i]);
35 int mx = 0 ;
36 for(int j = 1;j <= numa ;j ++)
37 {
38 if(b[i] > a[j])
39 {
40 mx = max(mx,dp[j]);
41 }else if(b[i] == a[j]){
42 dp[j] = mx+1;
43 }
44 }
45 }
46 int mx = 0 ;
47 for(int i= 1;i <= numa;i ++)
48 {
49 mx = max(dp[i],mx);
50 }
51 printf("%d\n",mx);
52 }
53 return 0 ;
54 }
时间： 2024-08-26 23:10:16