# legend_noa's blog

LFYZ蒟蒻竟然写题解！！！……

## 洛谷10月月赛II题解

T1找性质，popcnt为1的加上popcnt为0的就是答案（然而当时没想到只打了暴力）

#include<bits/stdc++.h>
typedef long long ll;
ll a, b, c, d, n, x;
int cnt[65536], sz[2];
int main(){
scanf("%d%lld%lld%lld%lld%lld", &n, &a, &b, &c, &d, &x);
for(int i = 0; i < 65536; i++) cnt[i] = cnt[i >> 1]^(i&1);
for(int i = 1; i <= n; i++){
x = (a*x%d*x%d+b*x%d+c)%d;
sz[cnt[x & 65535] ^ cnt[x >> 16]]++;
}
printf("%lld\n", (long long)sz[0]*sz[1]);
return 0;
}


T2枚举公差和塔即可

#include<bits/stdc++.h>
const int M = 998244353;
inline int chk(int x){return x >= M? x-M : x;}
int n, max, ans=1, h[1000001];
int d[1001][40001];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &h[i]), max = std::max(max, h[i]);
for(int i = 2; i <= n; i++){
for(int j = 1; j < i; j++){
int num = d[j][h[i] - h[j] + 20001]+1;
d[i][h[i]-h[j]+20001] += num;
if(d[i][h[i]-h[j]+20001]>=M) d[i][h[i]-h[j]+20001] -= M;
}
for(int j = -max+20000; j <= max+20000; j++) ans = chk(ans+d[i][j]);
ans = chk(ans+1);
}
printf("%d\n", ans);
return 0;
}


T3是一个DAG上的最长路问题，看懂题解后发现真**简单

#include<bits/stdc++.h>
const int N = 2e6+4;
int n, k, s, f[N]; std::vector<int> ans[21];int a[N];
bool vis[N];
int main(){
scanf("%d%d", &n, &k); s = (1 << k) -1;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]), vis[a[i]] = true;
for(int i = s; i>=0; i--){
if(vis[i]) f[i]++;
for(int j = 1; j < s; j <<= 1)
if(i&j) f[i^j] = std::max(f[i^j], f[i]);
}
printf("1\n%d\n", f[0]);
for(int i = s; i>=0; i--) if(vis[i]) ans[f[i]].push_back(i);
for(int i = 1; i <= f[0]; i++){
int size = ans[i].size(); printf("%d ", size);
for(int j = 0; j < size; j++) printf("%d ", ans[i][j]);
puts("");
}
return 0;
}


T4是一道自始至终都没看懂的数论题，只好先拿租酥雨大佬的代码填填坑

#include<bits/stdc++.h>
const int mod = 1000000007;using std::map; using std::pair;
inline void inc(int &x,int y){x+=y;x>=mod?x-=mod:x;}
inline void dec(int &x,int y){x-=y;x<0?x+=mod:x;}
int fastpow(int a,int b){
int res=1;
while (b) {if (b&1) res=1ll*res*a%mod;a=1ll*a*a%mod;b>>=1;}
return res;
}
int n,P,R,B,sqr,S,inv[5005],C[505][505],ans;
struct poly{
int a[505];
poly(){memset(a,0,sizeof(a));}
poly operator * (poly b){
poly c;
for (int i=0;i<=n;++i)
for (int j=0;j<=i;++j)
c.a[i]=(c.a[i]+1ll*a[j]*b.a[i-j]%mod*C[i][j])%mod;
return c;
}
}dp[3][2][80],zero;
map<pair<pair<int,int>,int>,int>M;
int calc(int u,int v,int w){
return (dp[0][0][u%sqr]*dp[0][1][u/sqr]*dp[1][0][v%sqr]*dp[1][1][v/sqr]*dp[2][0][w%sqr]*dp[2][1][w/sqr]*zero).a[n];
}
int main(){
scanf("%d%d%d", &n, &P, &R);
B=R/P;while (sqr*sqr<P) ++sqr;
inv[1]=1;for (int i=2;i<P;++i) inv[i]=inv[P%i]*(P-P/i)%P;
for (int i=C[0][0]=1;i<=n;++i)
for (int j=C[i][0]=1;j<=i;++j)
C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
for (int i=0;i<3;++i){
dp[i][0][0].a[0]=dp[i][1][0].a[0]=dp[i][0][1].a[0]=1;
for (int j=1;j<=n;++j) dp[i][0][1].a[j]=(fastpow(B+(i>0),j)+fastpow(B+(i>1),j))%mod;
for (int j=2;j<=sqr;++j) dp[i][0][j]=dp[i][0][j-1]*dp[i][0][1];
for (int j=1;j<=sqr;++j) dp[i][1][j]=dp[i][1][j-1]*dp[i][0][sqr];
}
for (int i=0;i<=n;++i) zero.a[i]=fastpow(B,i);
S=fastpow(R,n);inc(ans,S);dec(ans,fastpow(R-B,n));
for (int i=1;i<P;++i){    int u=0,v=0,w=0;
for (int j=1;j<P;++j){
int x=inv[j]*i%P;
if (x<j){ if (j<=R-B*P) ++w;  else if (x<=R-B*P) ++v;  else ++u;}
}
pair<pair<int,int>,int>pr=std::make_pair(std::make_pair(u,v),w);
if (M.find(pr)==M.end()) M[pr]=calc(u,v,w);
inc(ans,S);dec(ans,M[pr]);
}
printf("%d\n",ans);
return 0;
}


posted on 2018-10-24 18:22  legend_noa  阅读(91)  评论(0编辑  收藏  举报