模拟3
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ID
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Origin
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Title
| ||
|---|---|---|---|---|
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6 / 12 | Problem A | ZOJ 3860 | Find the Spy |
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6 / 27 | Problem B | ZOJ 3861 | Valid Pattern Lock |
| Problem C | ZOJ 3862 | Intersection | ||
| 0 / 1 | Problem D | ZOJ 3863 | Paths on the Tree | |
| Problem E | ZOJ 3864 | Quiz for EXO-L | ||
| Problem F | ZOJ 3865 | Superbot | ||
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5 / 15 | Problem G | ZOJ 3866 | Cylinder Candy |
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7 / 15 | Problem H | ZOJ 3867 | Earthstone: Easy Version |
| 1 / 6 | Problem I | ZOJ 3868 | GCD Expectation |
A.
d.找出不同的数。。
#include<iostream> #include<stdio.h> using namespace std; int main(){ int T; int N; int ID[128]; int i; int sum1,sum2; int p1,p2; scanf("%d",&T); while(T--){ scanf("%d",&N); for(i=0;i<N;++i){ scanf("%d",&ID[i]); } p1=0; for(i=1;i<N;++i){ if(ID[i]!=ID[i-1]){ p2=i; break; } } sum1=0; sum2=0; for(i=0;i<N;++i){ if(ID[i]==ID[p1]){ ++sum1; } else if(ID[i]==ID[p2]){ ++sum2; } } if(sum1==1){ printf("%d\n",ID[p1]); } else{ printf("%d\n",ID[p2]); } } return 0; }
B.
d.手机滑屏解锁的问题，给出几个1~9之间的几个数，求出密码的所有可能。
s.dfs
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; bool exist[16]; bool vis[16]; int can[16][16]; int t_path[212345][10]; int path[16]; int n; int sum; void dfs(int u,int now){ int i; if(now>n){ for(i=1;i<=9;++i){ t_path[sum][i]=path[i]; } ++sum; return; } for(i=1;i<=9;++i){ if(exist[i]&&!vis[i]){ if( (u==1&&(i==3||i==9||i==7)) || (u==3&&(i==1||i==7||i==9)) || (u==9&&(i==7||i==1||i==3)) || (u==7&&(i==9||i==3||i==1)) ){ if(vis[can[u][i]]){ path[now]=i; vis[i]=true; dfs(i,now+1); vis[i]=false; } else{ } } else if( (u==2&&i==8) || (u==8&&i==2) || (u==4&&i==6) || (u==6&&i==4) ){ if(vis[can[u][i]]){ path[now]=i; vis[i]=true; dfs(i,now+1); vis[i]=false; } else{ } } else{ path[now]=i; vis[i]=true; dfs(i,now+1); vis[i]=false; } } } } void init(){ memset(exist,false,sizeof(exist)); } int main(){ memset(can,0,sizeof(can)); can[1][3]=2; can[1][9]=5; can[1][7]=4; can[3][1]=2; can[3][7]=5; can[3][9]=6; can[9][7]=8; can[9][1]=5; can[9][3]=6; can[7][9]=8; can[7][3]=5; can[7][1]=4; //--- can[2][8]=5; can[8][2]=5; can[4][6]=5; can[6][4]=5; int T; int i; int t; int j; scanf("%d",&T); while(T--){ scanf("%d",&n); init(); for(i=0;i<n;++i){ scanf("%d",&t); exist[t]=true; } sum=0; for(i=1;i<=9;++i){ memset(vis,false,sizeof(vis)); if(exist[i]){ path[1]=i; vis[i]=true; dfs(i,2); vis[i]=false; } } printf("%d\n",sum); for(i=0;i<sum;++i){ for(j=1;j<n;++j){ printf("%d ",t_path[i][j]); } printf("%d\n",t_path[i][n]); } } return 0; }
C.
d.线段相交拆分问题
s.先排序，好像还可以。。
D.
E.
d.黑白图像的问题
ps:好像用连通块来判断，还有黑白点的比例。题目比较长，没读题。
F.
G.
d.积分求体积，求面积。
#include <iostream> #include <cmath> #include <cstdio> using namespace std; #define PI M_PI int main() { int T; double r,h,d,v,s; scanf("%d",&T); while(T--) { scanf("%lf%lf%lf",&r,&h,&d); v=h*PI*(r+d)*(r+d)+PI*r*r*d*2+2*2*PI*(1.0/3*d*d*d+d*d/4*PI*r); s=2*(PI*r*r+PI*(r+d)*h+2*PI*d*d+PI*PI*r*d); printf("%.12f %.12f\n",v,s); } return 0; }
H.
d.炉石传说也来水题了。。
#include<iostream> #include<stdio.h> using namespace std; int main(){ int T; int A1,H1; int A2,H2; int H3,H4; scanf("%d",&T); while(T--){ scanf("%d%d%d%d",&A1,&H1,&A2,&H2); if(A1==0){ printf("Invalid\n"); continue; } H3=H1-A2; H4=H2-A1; if(H3<=0){ printf("Discard "); } else{ printf("%d %d ",A1,H3); } if(H4<=0){ printf("Discard\n"); } else{ printf("%d %d\n",A2,H4); } } return 0; }
I.
ps:又是最大公约数的问题，尴尬。。
d.给出n和k, 求出n个数的任意非空子集的最大公约数的k次方的期望, 最后求出期望*(2^n-1)
s.容斥原理
每取一个子集的概率都为 1/(2^n-1), 结果除以(2^n-1),则实际上是求出每个非空子集的k次方的和.
题目转化为gcd为i的子集为多少, 枚举i 从MAX到1.
对于每个i, 求出n个数中为i 的倍数为cnt个,
则gcd=i 的非空子集的个数dp[i] = (2^cnt-1) - dp[j] (j为i 的倍数).
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; typedef long long ll; const int maxn = 1e6+5; const ll mod = 998244353; int sum[maxn]; ll dp[maxn]; ll cal(ll x, int n) { ll s = 1; while(n > 0) { if(n & 1) s = (s*x) % mod; x = (x*x)%mod; n >>= 1; } return s; } int main() { //freopen("in", "r", stdin); int T; scanf("%d", &T); while(T--) { int n, k; scanf("%d %d", &n, &k); memset(sum, 0, sizeof(sum)); memset(dp, 0, sizeof(dp)); int MAX = 0; for(int i = 0; i < n; ++i) { int a; scanf("%d", &a); MAX = max(MAX, a); sum[a]++; } ll ans = 0; for(int i = MAX; i >= 1; --i) { int cnt = 0; for(int j = i; j <= MAX; j+=i) { cnt += sum[j]; dp[i] = (dp[i] - dp[j] + mod) % mod; } dp[i] = ((dp[i]+cal(2, cnt)-1)%mod + mod) % mod; ans = (ans + dp[i]*cal(i, k)) % mod; } printf("%lld\n", ans); } return 0; }
c2.上个莫比乌斯反演的代码，还不懂
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <set> #include <queue> #include <vector> #define LL long long #define MOD 998244353 using namespace std; const int maxN = 1e6+10; int n, k, len; int pow2[maxN], num[maxN]; int prime[maxN], u[maxN]; bool vis[maxN]; //莫比乌斯反演 //F(n)和f(n)是定义在非负整数集合上的两个函数 //并且满足条件F(n) = sum(f(d)) (d|n) //那么f(n) = sum(u(d)*F(n/d)) (d|n) //case 1: if d = 1, u(d) = 1 //case 2: if d = p1*p2...pk, (pi均为互异素数), u(d) = (-1)^k //case 3: else, u(d) = 0; //性质1: sum(u(d)) (d|n) = 1 (n == 1) or 0 (n > 1) //性质2: sum(u(d)/d) (d|n) = phi(n)/n //另一种形式:F(d) = sum(f(n)) (d|n) => f(d) = sum(u(n/d)*F(n)) (d|n) //线性筛选求莫比乌斯反演函数代码 void mobius() { memset(vis, false,sizeof(vis)); u[1] = 1; int cnt = 0; for(int i = 2; i < maxN; i++) { if(!vis[i]) { prime[cnt++] = i; u[i] = -1; } for(int j = 0; j < cnt && i*prime[j] < maxN; j++) { vis[i*prime[j]] = true; if(i%prime[j]) u[i*prime[j]] = -u[i]; else { u[i*prime[j]] = 0; break; } } } } void init() { mobius(); pow2[0] = 1; for (int i = 1; i < maxN; ++i) pow2[i] = 2*pow2[i-1]%MOD; } void input() { int tmp; len = 0; scanf("%d%d", &n, &k); memset(num, 0, sizeof(num)); for (int i = 0; i < n; ++i) { scanf("%d", &tmp); len = max(len, tmp); num[tmp]++; } for (int i = 1; i <= len; ++i) { for (int j = i*2; j <= len; j += i) num[i] += num[j]; } } //快速幂m^n LL quickPow(LL x, int n) { if (n == 0) return 1; if (x == 0) return 0; LL a = 1; while (n) { a *= n&1 ? x : 1; a %= MOD; n >>= 1 ; x *= x; x %= MOD; } return a; } void work() { int ans = 0, val; for (int i = 1; i <= len; ++i) { val = quickPow(i, k); for (int j = i; j <= len; j += i) { ans += ((LL)u[j/i]*val%MOD)*(pow2[num[j]]-1)%MOD; ans = (ans%MOD+MOD)%MOD; } } printf("%d\n", ans); } int main() { //freopen("test.in", "r", stdin); init(); int T; scanf("%d", &T); for (int times = 1; times <= T; ++times) { input(); work(); } return 0; }
