[Offer收割]编程练习赛57

A.1-偏差排列

dp？

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 LL f[55][3];
 5
 6 int main(){
 7     int N;
 8     scanf("%d", &N);
 9     f[1][0] = f[1][2] = 1;
10     for(int i = 1; i < N; ++i){
11         f[i+1][0] = f[i][0] + f[i][1];
12         f[i+1][1] = f[i][2];
13         f[i+1][2] = f[i][0] + f[i][1];
14     }
15     printf("%lld\n", f[N][0] + f[N][1]);
16     return 0;
17 }

Aguin

B.递增N元组

dp？

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL mod = 1e9 + 7;
 5 LL f[105][1000005];
 6
 7 int main(){
 8     int N, M, x;
 9     scanf("%d %d", &N, &M);
10     for(int i = 0; i <= 100001; ++i) f[0][i] = 1;
11     for(int i = 1; i <= N; ++i){
12         for(int j = 1; j <= M; ++j) {
13             scanf("%d", &x), x++;
14             f[i][x] = (f[i][x] + f[i-1][x-1]) % mod;
15         }
16         for(int j = 1; j <= 100001; ++j) f[i][j] = (f[i][j] + f[i][j-1]) % mod;
17     }
18     printf("%lld\n", f[N][100001]);
19     return 0;
20 }

Aguin

C.逃离迷宫5

bfs？

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 typedef pair<pii, int> piii;
 5 const int INF = 1e9;
 6 int G[1111][1111], f[1111][1111][2];
 7 int dx[] = {-1, 0, 1, 0};
 8 int dy[] = {0, 1, 0, -1};
 9 char s[1111];
10 queue<piii> q;
11
12 int main(){
13     int N;
14     scanf("%d", &N);
15     for(int i = 1; i <= N; ++i){
16         scanf("%s", s + 1);
17         for(int j = 1; j <= N; ++j) G[i][j] = s[j] == ‘#‘ ? 1 : 0, f[i][j][0] = f[i][j][1] = INF;
18     }
19     if(G[1][1]) q.push(piii(pii(1, 1), 1)), f[1][1][1] = 0;
20     else q.push(piii(pii(1, 1), 0)), f[1][1][0] = 0;
21     while(!q.empty()){
22         piii now = q.front(); q.pop();
23         int nx = now.first.first, ny = now.first.second, o = now.second;
24         for(int i = 0; i < 4; ++i){
25             int xx = nx + dx[i], yy = ny + dy[i];
26             if(xx < 1 || xx > N || yy < 1 || yy > N) continue;
27             if(G[xx][yy] && o) continue;
28             int oo = o || G[xx][yy];
29             if(f[xx][yy][oo] != INF) continue;
30             f[xx][yy][oo] = f[nx][ny][o] + 1;
31             q.push(piii(pii(xx, yy), oo));
32         }
33     }
34     int ans = min(f[N][N][0], f[N][N][1]);
35     printf("%d\n", ans == INF ? -1 : ans);
36     return 0;
37 }

Aguin

D.最大割集

……无语了

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long LL;
  4 const int maxn = 1e5 + 10;
  5 LL W[maxn], V[maxn], b[66];
  6 int id[maxn];
  7
  8 bool cmp(int i, int j){
  9     return W[i] > W[j];
 10 }
 11
 12 // Template
 13 //清除前缀0，如果结果是空字符串则设为0
 14 inline void clear(string& a){
 15     while(a.length()>0 && a[0]==‘0‘)
 16         a.erase(0, 1);
 17     if(a == "")
 18         a = "0";
 19 }
 20
 21 //如果a>=b则返回真（如果包含前缀零会被消除）
 22 bool isBigger(string a, string b){
 23     clear(a);
 24     clear(b);
 25     if(a.length() > b.length())
 26         return true;
 27     if(a.length()==b.length() && a>=b)
 28         return true;
 29     return false;
 30 }
 31
 32
 33 //两个高精度正整数加法 a+b
 34 string Add(string a, string b){
 35     //1、对位，将两个数补零直到其具有相同长度
 36     while(a.length() < b.length())
 37         a = ‘0‘ + a;
 38     while(a.length() > b.length())
 39         b = ‘0‘ + b;
 40     //2、补零，在开头再加一个0以便进位
 41     a = ‘0‘ + a;
 42     b = ‘0‘ + b;
 43     //3、从低位开始相加，注意进位
 44     for(int i=a.length()-1; i>=0; i--){
 45         a[i] = a[i] + b[i] - ‘0‘;
 46         if(a[i] > ‘9‘){
 47             a[i] = a[i] - 10;
 48             a[i-1] += 1;
 49         }
 50     }
 51     clear(a);
 52     return a;
 53 }
 54
 55 //两个高精度正整数减法 a-b
 56 string Minus(string a, string b){
 57     bool aBigger = true;
 58     //1、对位，将两个数补零直到其具有相同长度
 59     while(a.length() < b.length())
 60         a = ‘0‘ + a;
 61     while(a.length() > b.length())
 62         b = ‘0‘ + b;
 63     //2、推测结果正负值，调整为前大后小
 64     if(a < b)
 65     {
 66         aBigger = false;
 67         string buf = b;
 68         b = a;
 69         a = buf;
 70     }
 71     //3、从低位开始相减，注意借位
 72     for(int i=a.length()-1; i>=0; i--){
 73         if(a[i] >= b[i]){
 74             a[i] = a[i] - (b[i] - ‘0‘);
 75         }else{
 76             a[i] = a[i] + 10;
 77             a[i-1] -= 1;
 78             a[i] = a[i] - (b[i] - ‘0‘);
 79         }
 80     }
 81     clear(a);
 82     if(!aBigger)
 83         a = ‘-‘ + a;
 84     return a;
 85 }
 86
 87 template<typename T> string to(const T& t){
 88     ostringstream oss;  //创建一个格式化输出流
 89     oss<<t;             //把值传递如流中
 90     return oss.str();
 91 }
 92
 93 int main(){
 94     int N, M;
 95     scanf("%d %d", &N, &M);
 96     string sum("0"), ans("0");
 97     for(int i = 1; i <= N; ++i) scanf("%lld", W + i), sum = Add(sum, to(W[i])), id[i] = i;
 98     sort(id + 1, id + 1 + N, cmp);
 99     for(int i = 1; i <= M; ++i){
100         int u, v;
101         LL w;
102         scanf("%d %d %lld", &u, &v, &w);
103         V[u] ^= w, V[v] ^= w;
104     }
105     for(int i = 1; i <= N; ++i){
106         int x = id[i];
107         for(int j = 0; j <= 60; ++j) {
108             if(V[x] & (1LL << j)){
109                 if(b[j]) V[x] ^= b[j];
110                 else {b[j] = V[x], ans = Add(ans, to(W[x])); break;}
111             }
112         }
113     }
114     cout << Minus(Add(ans, ans), sum) << endl;
115     return 0;
116 }

Aguin

学习了py

 1 N, M = list(map(int, raw_input().strip().split()))
 2 W = list(map(int, raw_input().strip().split()))
 3 sum = sum(W)
 4 id = sorted([(W[i], i) for i in range(N)], reverse=True)
 5 V = [0 for i in range(N)]
 6 for i in range(M):
 7     u, v, w = list(map(int, raw_input().strip().split()))
 8     V[u-1] ^= w
 9     V[v-1] ^= w
10 b = [0 for i in range(61)]
11 ans = 0
12 for i in range(N):
13     x = id[i][1]
14     for j in range(61):
15         if V[x] & (1 << j):
16             if b[j]:
17                 V[x] ^= b[j]
18             else:
19                 b[j] = V[x]
20                 ans += W[x]
21                 break
22 print(ans + ans - sum)

Aguin

原文地址：https://www.cnblogs.com/Aguin/p/8971730.html