2018/02/26 模拟赛

第一题排序暴力

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<string>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<utility>
 8 #include<stdio.h>
 9 #include<cstdlib>
10 #include<iomanip>    //cout<<setiosflags(ios::fixed)<<setprecision(2);
11 #include<ctime> //double a=(double)clock();    cout<<a<<endl;
12 #include<vector>
13 #include<queue>
14 #include<set>
15 #include<map>
16 using namespace std;
17 int read(){
18     int x=0,ff=1;char ch=getchar();
19     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)ff=-1; ch=getchar();}
20     while(ch<=‘9‘&&ch>=‘0‘){x=x*10+ch-‘0‘;ch=getchar();}
21     return ff*x;
22 }
23 map<int,int>mp;
24 int N,a,b[100010];
25 long long ans;
26 int main(){
27     freopen("expedition.in","r",stdin);
28     freopen("expedition.out","w",stdout);
29     N=read();
30     for(int i=1;i<=N;i++){
31         a=read(),b[i]=read();
32         mp[b[i]]+=a;
33     }
34     sort(b+1,b+1+N);
35     for(int i=1;i<=N;i++)
36         if(b[i]!=b[i-1])
37             ans+=1LL*mp[b[i]]*mp[b[i]];
38     printf("%I64d\n",ans);
39     return 0;
40 }

第二题计算表达式，对于多项式用类似高精度的方法计算

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<string>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<utility>
  8 #include<stdio.h>
  9 #include<cstdlib>
 10 #include<iomanip>    //cout<<setiosflags(ios::fixed)<<setprecision(2);
 11 #include<ctime> //double a=(double)clock();    cout<<a<<endl;
 12 #include<vector>
 13 #include<queue>
 14 using namespace std;
 15 int read(){
 16     int x=0,ff=1;char ch=getchar();
 17     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)ff=-1; ch=getchar();}
 18     while(ch<=‘9‘&&ch>=‘0‘){x=x*10+ch-‘0‘;ch=getchar();}
 19     return ff*x;
 20 }
 21 const int MOD=10007;
 22 struct bignum{
 23     int num[510];
 24     int len;
 25     void clear()
 26     {len=0,memset(num,0,sizeof(num));}
 27     void print(){
 28         printf("%d\n",len);
 29         for(int i=0;i<=len;i++){
 30             if(num[i]<0)
 31                 num[i]+=MOD;
 32             printf("%d\n",num[i]);
 33         }
 34     }
 35     bignum friend operator+(const bignum&A,const bignum&B){
 36         bignum C;
 37         C.len=max(A.len,B.len);
 38         for(int i=0;i<=C.len;i++){
 39             if(i>A.len)
 40                 C.num[i]=B.num[i];
 41             else if(i>B.len)
 42                 C.num[i]=A.num[i];
 43             else
 44                 C.num[i]=(A.num[i]+B.num[i])%MOD;
 45         }
 46         while(C.num[C.len]==0&&C.len)
 47             C.len--;
 48         return C;
 49     }
 50     bignum friend operator-(const bignum&A,const bignum&B){
 51         bignum C;
 52         C.len=max(A.len,B.len);
 53         for(int i=0;i<=C.len;i++){
 54             if(i>A.len)
 55                 C.num[i]=-B.num[i];
 56             else if(i>B.len)
 57                 C.num[i]=A.num[i];
 58             else
 59                 C.num[i]=(A.num[i]-B.num[i])%MOD;
 60         }
 61         while(C.num[C.len]==0&&C.len)
 62             C.len--;
 63         return C;
 64     }
 65     bignum friend operator*(const bignum&A,const bignum&B){
 66         bignum C;
 67         C.clear();
 68         for(int i=0;i<=A.len;i++)
 69             for(int j=0;j<=B.len;j++){
 70                 int k=i+j;
 71                 C.num[k]+=A.num[i]*B.num[j];
 72                 C.num[k]%=MOD;
 73                 if(k>C.len&&C.num[k]!=0)
 74                     C.len=k;
 75             }
 76         return C;
 77     }
 78 }p[1010],t;//多项式栈
 79 char s[1010];
 80 int N;
 81 char op[1010];//operator 运算符栈
 82 int to,tp;
 83 inline int grade(char c){
 84     if(c==‘+‘||c==‘-‘)
 85         return 1;
 86     else if(c==‘*‘)
 87         return 2;
 88     else
 89         return 0;
 90 }
 91 int main(){
 92     freopen("simplify.in","r",stdin);
 93     freopen("simplify.out","w",stdout);
 94     gets(s+1);
 95     N=strlen(s+1);
 96     for(int i=1;i<=N;i++){
 97         if(s[i]==‘x‘){
 98             tp++;
 99             p[tp].len=1;
100             p[tp].num[0]=0;
101             p[tp].num[1]=1;
102         }
103         else if(s[i]>=‘0‘&&s[i]<=‘9‘){
104             int j=i,temp=0;
105             while(s[j]>=‘0‘&&s[j]<=‘9‘){
106                 temp=temp*10+s[j]-‘0‘;
107                 j++;
108             }
109             tp++;
110             p[tp].len=0;
111             p[tp].num[0]=temp;
112             i=j-1;
113         }
114         else{
115             if(s[i]==‘(‘)
116                 op[++to]=s[i];
117             else if(s[i]==‘)‘){
118                 while(to&&op[to]!=‘(‘){
119                     tp--;
120                     if(op[to]==‘+‘)
121                         p[tp]=p[tp]+p[tp+1];
122                     else if(op[to]==‘-‘)
123                         p[tp]=p[tp]-p[tp+1];
124                     else
125                         p[tp]=p[tp]*p[tp+1];
126                     to--;
127                 }
128                 to--;
129             }
130             else{
131                 while(to&&grade(op[to])>=grade(s[i])){
132                     tp--;
133                     if(op[to]==‘+‘)
134                         p[tp]=p[tp]+p[tp+1];
135                     else if(op[to]==‘-‘)
136                         p[tp]=p[tp]-p[tp+1];
137                     else
138                         p[tp]=p[tp]*p[tp+1];
139                     to--;
140                 }
141                 op[++to]=s[i];
142             }
143         }
144     }
145     while(to){
146         tp--;
147         if(op[to]==‘+‘)
148             p[tp]=p[tp]+p[tp+1];
149         else if(op[to]==‘-‘)
150             p[tp]=p[tp]-p[tp+1];
151         else
152             p[tp]=p[tp]*p[tp+1];
153         to--;
154     }
155     p[tp].print();
156     return 0;
157 }

第三题最短路，“一到多”的最短路可以直接求，“多到一“的最短路可以将所有边反向后做最短路来求

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<string>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<utility>
 8 #include<stdio.h>
 9 #include<cstdlib>
10 #include<iomanip>    //cout<<setiosflags(ios::fixed)<<setprecision(2);
11 #include<ctime> //double a=(double)clock();    cout<<a<<endl;
12 #include<vector>
13 #include<queue>
14 using namespace std;
15 int read(){
16     int x=0,ff=1;char ch=getchar();
17     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)ff=-1; ch=getchar();}
18     while(ch<=‘9‘&&ch>=‘0‘){x=x*10+ch-‘0‘;ch=getchar();}
19     return ff*x;
20 }
21 const int maxn=3010,maxm=100010;
22 int N,M,Q;
23 struct Edge{
24     int x,y,v;
25 }E[maxm];
26 struct edge{
27     int y,next,v;
28 }e[maxm*2];
29 int lin[maxn],len;
30 inline void insert(int xx,int yy,int vv){
31     e[++len].next=lin[xx];
32     lin[xx]=len;
33     e[len].y=yy;
34     e[len].v=vv;
35 }
36 void build(bool sgn){
37     memset(lin,0,sizeof(lin));len=0;
38     if(!sgn)
39         for(int i=1;i<=M;i++)
40             insert(E[i].x,E[i].y,E[i].v);
41     else
42         for(int i=1;i<=M;i++)
43             insert(E[i].y,E[i].x,E[i].v);
44 }
45 int dis[2][maxn],q[2000010],head,tail;
46 bool vis[maxn];
47 void SPFA(int sgn){
48     head=0,tail=0;
49     q[head]=1;
50     vis[1]=1;
51     dis[sgn][1]=0;
52     for(;head<=tail;head++){
53         for(int i=lin[q[head]];i;i=e[i].next)
54             if(dis[sgn][q[head]]+e[i].v<dis[sgn][e[i].y]){
55                 dis[sgn][e[i].y]=dis[sgn][q[head]]+e[i].v;
56                 if(!vis[e[i].y]){
57                     vis[e[i].y]=1;
58                     q[++tail]=e[i].y;
59                 }
60             }
61         vis[q[head]]=0;
62     }
63 }
64 int main(){
65     freopen("production.in","r",stdin);
66     freopen("production.out","w",stdout);
67     N=read(),M=read();
68     for(int i=1;i<=M;i++)
69         E[i].x=read(),E[i].y=read(),E[i].v=read();
70     memset(dis,10,sizeof(dis));
71     build(0);
72     SPFA(0);
73     build(1);
74     SPFA(1);
75     Q=read();
76     for(int i=1;i<=Q;i++){
77         int tx=read(),ty=read();
78         if(dis[1][tx]+dis[0][ty]>=dis[0][0])//dis[0][0]==INF
79             printf("-1\n");
80         else
81             printf("%d\n",dis[1][tx]+dis[0][ty]);
82     }
83     return 0;
84 }

原文地址：https://www.cnblogs.com/lzhAFO/p/8472604.html