NIOP模拟17.10.13

太水，简述一下题意

 

T1

让你计算一个形如Σai * bi^ki

 

快速幂即可

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <ctime>
 7 #define min(a, b) ((a) < (b) ? (a) : (b))
 8 #define max(a, b) ((a) > (b) ? (a) : (b))
 9 
10 inline void swap(long long &x, long long &y)
11 {
12     long long tmp = x;x = y;y = tmp;
13 }
14 
15 inline void read(long long &x)
16 {
17     x = 0;char ch = getchar(), c = ch;
18     while(ch < '0' || ch > '9')c = ch, ch = getchar();
19     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
20     if(c == '-')x = -x;
21 }
22 
23 const long long INF = 0x3f3f3f3f;
24 const long long MAXN = 1000000;
25 
26 long long n,k,MOD,ans,x;
27 
28 long long pow(long long a, long long b)
29 {
30     long long r = 1, base = a;
31     for(;b;b >>= 1)
32     {
33         if(b & 1) r *= base, r %= MOD;
34         base *= base, base %= MOD;
35     }
36     return r;
37 }
38 
39 long long tmp[100000], tot, a[MAXN], b[MAXN];
40 
41 int main()
42 {
43     freopen("digits.in", "r", stdin);
44     freopen("digits.out", "w", stdout);
45     read(n), read(k);
46     MOD = 1;
47     for(register long long i = 1;i <= k;++ i) MOD *= 10;
48     for(register long long i = 1;i <= n;++ i)
49         read(a[i]), read(b[i]);
50     read(x);
51     for(register long long i = 1;i <= n;++ i)
52     {
53         ans += (pow(x, b[i]) * a[i])%MOD, ans %= MOD;
54     }
55     for(register long long i = 1;i <= k;++ i)
56     {
57         tmp[++tot] = ans % 10;
58         ans /= 10;
59     }
60     for(register long long i = k;i >= 1;-- i)
61         printf("%d\n", tmp[i]);
62     return 0;
63 }
T1

 

T2

让你求a1 * x1 - x2 * x2 + x3 * x3 - a4 * x4 + a5 * x5 - a6 * x6 = 0在(0,k]内的解的个数，ai <= 10^5, k <= 600

 

meet in the middle可过，map做

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <ctime>
 7 #include <map>
 8 #define min(a, b) ((a) < (b) ? (a) : (b))
 9 #define max(a, b) ((a) > (b) ? (a) : (b))
10 
11 inline void swap(long long &x, long long &y)
12 {
13     long long tmp = x;x = y;y = tmp;
14 }
15 
16 inline void read(long long &x)
17 {
18     x = 0;char ch = getchar(), c = ch;
19     while(ch < '0' || ch > '9')c = ch, ch = getchar();
20     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
21     if(c == '-')x = -x;
22 }
23 
24 const long long INF = 0x3f3f3f3f;
25 
26 std::map<long long, long long> mp;
27 long long num1[4], num2[4], k, ans;
28 
29 int main()
30 {
31     freopen("equation.in", "r", stdin);
32     freopen("equation.out", "w", stdout);
33     read(k);
34     for(register long long i = 1;i <= 6;++ i)
35     {
36         if(i & 1)read(num1[(i - 1)/2 + 1]);
37         else read(num2[(i - 1)/2 + 1]);
38     }
39     for(register long long i = 1;i <= k;++ i)
40         for(register long long j = 1;j <= k;++ j)
41             for(register long long q = 1;q <= k;++ q)
42                 ++ mp[i * num1[1] + j * num1[2] + q * num1[3]]; 
43     for(register long long i = 1;i <= k;++ i)
44         for(register long long j = 1;j <= k;++ j)
45             for(register long long q = 1;q <= k;++ q)
46                 if(mp.count(i * num2[1] + j * num2[2] + q * num2[3]))
47                     ans += mp[i * num2[1] + j * num2[2] + q * num2[3]];
48     printf("%lld\n", ans);
49     return 0;
50 }
T2

 

 

T3

让你求图上1到n的路径上的最大权值最小是多少

最小生成树 + dfs骚操作

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <algorithm>
 6 #define min(a, b) ((a) < (b) ? (a) : (b))
 7 #define max(a, b) ((a) > (b) ? (a) : (b))
 8 
 9 inline void swap(long long &x, long long &y)
10 {
11     long long tmp = x;x = y;y = tmp;
12 }
13 
14 inline void read(long long &x)
15 {
16     x = 0;char ch = getchar(), c = ch;
17     while(ch < '0' || ch > '9')c = ch, ch = getchar();
18     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
19     if(c == '-')x = -x;
20 }
21 
22 const long long INF = 0x3f3f3f3f;
23 const long long MAXN = 1000000 + 10;
24 
25 long long u[MAXN], v[MAXN], w[MAXN], cnt[MAXN];
26 
27 struct Edge
28 {
29     long long u,v,w,next;
30     Edge(long long _u, long long _v, long long _next, long long _w){w = _w;u = _u, v = _v,next = _next;}
31     Edge(){}
32 }edge[MAXN << 1];
33 long long head[MAXN], cntt;
34 
35 inline void insert(long long a, long long b, long long c)
36 {
37     edge[++cntt] = Edge(a,b,head[a],c);
38     head[a] = cntt;
39 }
40 
41 long long n,m,k,fa[MAXN];
42 
43 long long find(long long x)
44 {
45     return fa[x] == x ? x : fa[x] = find(fa[x]);
46 }
47 
48 long long cmp(long long a, long long b)
49 {
50     return w[a] < w[b];
51 }
52 
53 long long ans, b[MAXN];
54 
55 long long dfs(long long u)
56 {
57     b[u] = 1;
58     for(register long long pos = head[u];pos;pos = edge[pos].next)
59     {
60         long long v = edge[pos].v;
61         if(v == n)
62         {
63             ans = max(ans, edge[pos].w);
64             return 1;
65         }
66         if(b[v]) continue;
67         if(dfs(v))
68         {
69             ans = max(ans, edge[pos].w);
70             return 1;
71         }
72     }
73     return 0;
74 }
75 
76 int main()
77 {
78     freopen("graph.in", "r", stdin);
79     freopen("graph.out", "w", stdout);
80     read(n), read(m), read(k);
81     for(register long long i = 1;i <= m;++ i)
82         read(u[i]), read(v[i]), read(w[i]), cnt[i] = i;
83     std::sort(cnt + 1, cnt + 1 + m, cmp);
84     for(register long long i = 1;i <= n;++ i) fa[i] = i;
85     for(register long long p = 1;p <= m;++ p)
86     {
87         long long i = cnt[p];
88         long long f1 = find(u[i]), f2 = find(v[i]);
89         if(f1 == f2)continue;
90         fa[f1] = f2;
91         insert(u[i], v[i], w[i]);
92         insert(v[i], u[i], w[i]);
93     }
94     dfs(1);
95     if(ans == 0)ans = -1;
96     printf("%lld", ans);
97     return 0;
98 }
T3

 

posted @ 2017-10-13 17:05  嘒彼小星  阅读(334)  评论(0)  编辑  收藏  举报