转圈游戏
题解：快速幂
1 #include <cstdio>
2
3 int n, m, k, x;
4
5 inline long long QuickPow(int a, int k, int MOD){
6 long long base = a, ret = 1;
7 while (k){
8 if (k&1) ret = (ret*base)%MOD;
9 base = (base*base)%MOD, k >>= 1;
10 }
11 return ret;
12 }
13
14 int main(){
15 scanf("%d %d %d %d", &n, &m, &k, &x);
16 printf("%lld\n", (x%n+(m%n)*QuickPow(10, k, n)%n)%n);
17 }
circle.cpp
火柴排队
题解：求逆序对(因为写不来归排就写的树状数组，怕TLE就蛋疼的加了一个读入优化)
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5
6 const int MAXN = 100000+10;
7 const int MOD = 99999997;
8
9 int n, ans, c[MAXN], d[MAXN];
10 char cc;
11
12 struct Match{
13 int v, n;
14
15 friend bool operator < (const Match& A, const Match& B){
16 return A.v<B.v;
17 }
18 }a[MAXN], b[MAXN];
19
20 inline void add(int p, int v){
21 while (p<=n)
22 d[p] += v, p += (p&(-p));
23 }
24
25 inline int sum(int p){
26 int ret = 0;
27 while (p>0)
28 ret += d[p], p -= (p&(-p));
29 return ret;
30 }
31
32 inline int NextInt(){
33 int ret = 0;
34 do cc = getchar();
35 while (cc<48 || cc>57);
36 do ret *= 10, ret += (cc-48), cc = getchar();
37 while (48<=cc && cc<=57);
38 return ret;
39 }
40
41 int main(){
42 memset(d, 0, sizeof(0));
43 n = NextInt(), ans = 0;
44 for (int i=0; i<n; i++)
45 a[i].v = NextInt(), a[i].n = i+1;
46 sort(a, a+n);
47 for (int i=0; i<n; i++)
48 b[i].v = NextInt(), b[i].n = i+1;
49 sort(b, b+n);
50
51 for (int i=0; i<n; i++)
52 c[a[i].n] = b[i].n;
53
54 for (int i=1; i<=n; i++)
55 add(c[i], 1),
56 ans = (ans+i-sum(c[i]))%MOD;
57 printf("%d\n", ans);
58 }
match.cpp
货车运输
题解：先建一个最大生成树，再用lca乱搞一下就好了
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5
6 const int MAXL = 20;
7 const int MAXN = 10000+10;
8 const int MAXM = 50000+10;
9 const int INF = 0x3f3f3f3f;
10
11 int n, m, f[MAXN], nimo[MAXN][MAXL], an[MAXN][MAXL], dep[MAXN];
12 bool vis[MAXN];
13 char c;
14
15 struct Edge{
16 int d, to;
17 Edge *next;
18 }e[MAXM], *head[MAXN];
19
20 struct Bary{
21 int u, v, d;
22
23 friend bool operator < (const Bary& A, const Bary& B){
24 return A.d>B.d;
25 }
26 }bd[MAXM];
27
28 inline int NextInt(){
29 int ret = 0;
30 do c = getchar();
31 while (c<48 || c>57);
32 do ret *= 10, ret += (c-48), c = getchar();
33 while (48<=c && c<=57);
34 return ret;
35 }
36
37 inline int find(int x){
38 return x==f[x] ? x : f[x]=find(f[x]);
39 }
40
41 int ne = 0;
42 inline void AddEdge(int f, int t, int d){
43 e[ne].to = t, e[ne].d = d;
44 e[ne].next = head[f];
45 head[f] = e + ne++;
46 }
47
48 inline void Add(int u, int v, int d){
49 f[find(u)] = find(v);
50 AddEdge(u, v, d), AddEdge(v, u, d);
51 }
52
53 inline void BuildTree(int x){
54 vis[x] = true;
55 for (Edge *p=head[x]; p; p=p->next)
56 if (!vis[p->to])
57 an[p->to][0] = x, nimo[p->to][0] = p->d, dep[p->to] = dep[x]+1, BuildTree(p->to);
58 }
59
60 inline int swim(int &x, int ht){
61 int ret = INF;
62 for (int i=MAXL-1; i>=0; i--)
63 if (dep[an[x][i]]>=ht) ret = min(ret, nimo[x][i]), x = an[x][i];
64 return ret;
65 }
66
67 inline int lca(int x, int y){
68 int ret = INF;
69 if (dep[x]^dep[y]) ret = dep[x]>dep[y] ? swim(x, dep[y]) : swim(y, dep[x]);
70 if (x==y) return ret;
71 for (int i=MAXL-1; i>=0; i--)
72 if (an[x][i]^an[y][i])
73 ret = min(ret, min(nimo[x][i], nimo[y][i])),
74 x = an[x][i], y = an[y][i];
75 return min(ret, min(nimo[x][0], nimo[y][0]));
76 }
77
78 int main(){
79 memset(vis, false, sizeof(vis));
80 memset(an, 0, sizeof(an));
81
82 n = NextInt(), m = NextInt();
83 for (int i=1; i<=n; i++)
84 f[i] = i;
85 for (int i=0; i<m; i++)
86 bd[i].v = NextInt(), bd[i].u = NextInt(), bd[i].d = NextInt();
87
88 sort(bd, bd+m);
89 for (int i=0; i<m; i++)
90 if (find(bd[i].u)^find(bd[i].v)) Add(bd[i].u, bd[i].v, bd[i].d);
91 for (int i=1; i<=n; i++)
92 if (!vis[i]) dep[i] = 0, BuildTree(i), nimo[i][0] = INF, an[i][0] = i;
93
94 for (int i=1; i<MAXL; i++)
95 for (int j=1; j<=n; j++)
96 an[j][i] = an[an[j][i-1]][i-1],
97 nimo[j][i] = min(nimo[j][i-1], nimo[an[j][i-1]][i-1]);
98
99 int Q, a, b;
100 scanf("%d", &Q);
101 while (Q--)
102 scanf("%d %d", &a, &b),
103 printf("%d\n", find(a)==find(b) ? lca(a, b) : -1);
104 }
truck.cpp
机模大赛：
题解：模拟，ans = ∑max{0, a[i]-a[i-1]}
1 #include <cstdio>
2
3 int num, h[100010];
4
5 int max(int a,int b){
6 return a>b ? a : b;
7 }
8
9 int main(){
10 scanf("%d",&num);
11 for (int i=0; i<num; i++)
12 scanf("%d", &h[i]);
13
14 long long ans = 0;
15 for (int i=0; i<num; i++)
16 ans += max(0, h[i]-h[i-1]);
17 printf("%lld",ans);
18 }
block.cpp
花匠：
题解：贪心，缩点，把满足的连这的一群点缩为一个
1 #include <cstdio>
2 const int MAXN = 1e6+10;
3
4 int n, Pri, cj, a[MAXN];
5
6 int main(){
7 scanf("%d", &n), Pri = 1, cj = 1221;
8 for (int i=0; i<n; i++)
9 scanf("%d", a+i);
10 for (int i=1; i<n; i++){
11 if (a[i]>a[i-1] && cj!=1) cj = 1, Pri ++;
12 if (a[i]<a[i-1] && cj!=0) cj = 0, Pri ++;
13 }
14 printf("%d", Pri);
15 }
flower.cpp
NOIP2013 题解
时间： 2024-10-31 13:30:00