毒枭模拟(多校冲刺) noip 11.08
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毒枭模拟 noip 11.08
依旧是多校冲刺,这次的难度却是到达了一个不可思议的地步
考场最高分只有\(140pts\),我也只有\(100pts\)
关于考试心态:
今天考的时候心态又炸了,原因是调第一题调不出来,很生气
然后就在第一题上放了将近三个小时
当时我以为我要倒数第一了,没想到后三个题这么难......
T1
二分**板子题, 也可以用三分,然而这两个都有精度问题
二分整数太长了,三分小数不太够,cnm......
AC_code
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fo(i,x,y) for(int i=(x);i<=(y);i++)
#define fu(i,x,y) for(int i=(x);i>=(y);i--)
const int inf=0x3f3f3f3f3f3f3f3f;
int q,a,b,c,d,x,ans;
bool comp(int a,int y,int z){
if(z<0)return true;
if(a<0||y<0)return true;
if(y==0)return z<0;
else return a>z/y;
}
int check(int s){
int fi=a+b*(s-1);
if(fi<c)return 0;
return (fi-c)/d+1;
}
int get(int s){
int fi=a*s+b*s*(s-1)/2;
if(s==0)fi=0;
int l=0,r=x-fi,mid;
while(l<r){
mid=l+r+1>>1;
int sum=c*mid+d*(mid-1)*mid/2;
if(comp(d,mid-1,x-c)||comp(c,mid,x)||comp(d,mid*(mid-1)/2,x)||sum>x-fi||sum<0)r=mid-1;
else l=mid;
}
return l;
}
signed main(){
freopen("money.in","r",stdin);
freopen("money.out","w",stdout);
scanf("%lld",&q);
while(q--){
scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&x);
int l=1,r=x,mid,s2,s3;
while(l<r){
mid=l+r+1>>1;
s2=a*mid+b*mid*(mid-1)/2;
int ck=check(mid);
s3=c*ck+d*ck*(ck-1)/2;if(ck==0)s3=0;
if(comp(c,ck,x)||comp(d,ck*(ck-1)/2,x)||s3<0||s3>x)r=mid-1;
else if(comp(a,mid,x)||comp(b,mid*(mid-1)/2,x)||s2<0||s2>x||s2+s3>x||s2+s3<0)r=mid-1;
else l=mid;
}ans=get(0);
if(a*l+b*l*(l-1)/2>0&&a*l+b*l*(l-1)/2<=x){
ans=max(ans,l+get(l));
if(l>=1&&a*(l-1)+b*(l-2)*(l-1)/2>0&&a*(l-1)+b*(l-2)*(l-1)/2<=x)ans=max(ans,l-1+get(l-1));
}
printf("%lld\n",ans);
}
return 0;
}
T2
**多项式,真不会!!
直接跳了
T3
考场上竟然没有写数位\(DP\),失算了
\(n\)比较大的时候直接构造就行了
AC_code
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fo(i,x,y) for(int i=(x);i<=(y);i++)
#define fu(i,x,y) for(int i=(x);i>=(y);i--)
const int inf=0x3f3f3f3f3f3f3f3f;
int n,mod,w,m,ans;
int ksm(int x,int y){
int ret=1;
while(y){
if(y&1)ret=ret*x%mod;
x=x*x%mod;y>>=1;
}return ret;
}
int dp[405][9005];
int dfs(int x,int y){
if(!y||!x)return 1;
if(~dp[x][y])return dp[x][y];
dp[x][y]=0;
fo(i,max(0ll,y-9*(x-1)),min(9ll,y)){
dp[x][y]=dp[x][y]+dfs(x-1,y-i);
}
return dp[x][y];
}
void get_ans(int x,int y,int lim){
if(!x||!y)return ;
int now=lim;
fo(i,max(0ll,y-9*(x-1)),min(9ll,y)){
int del=dfs(x-1,y-i);
if(now<=dfs(x-1,y-i)){
ans=(ans*10ll+i)%mod;
get_ans(x-1,y-i,now);
break;
}
now-=del;
}
}
signed main(){
freopen("number.in","r",stdin);
freopen("number.out","w",stdout);
memset(dp,-1,sizeof(dp));
scanf("%lld%lld",&n,&mod);
fo(t,1,min(n,13ll)){
w=t*t*t;m=t*t*t*t;
bool fl=false;ans=0;
fo(i,1,inf){
fo(j,max(1ll,w-9*(i-1)),min(9ll,w)){
int now=dfs(i-1,w-j);
if(m<now){
fl=true;ans=j;
// cout<<j<<endl;
get_ans(i-1,w-j,m);
printf("%lld",ans);
printf(" ");
break;
}
m-=now;
}
if(fl)break;
}
}
int bas=0,p=ksm(10,40000);
fo(i,1,40000)bas=(bas*10+9)%mod;
fo(t,14,n){
w=t*t*t;m=t*t*t*t;
int now=w%9+2,sb=w/9;
//cout<<now<<endl;
// cout<<sb*(sb+1)/2<<" "<<m<<endl;
ans=now;m-=1+sb;
if(ans>9)ans=(ans%9*10+9)%mod;
fo(i,1,sb/40000)ans=(ans*p+bas)%mod;
fo(i,1,sb%40000)ans=(ans*10ll+9)%mod;
if(now>9)sb++;
fo(i,1,sb){
if(m<=sb-i+1){
//cout<<i<<" "<<sb-i+1-m<<endl;
ans=(ans-ksm(10,sb-i)+mod)%mod;
ans=(ans-ksm(10,sb-i+1-m)+mod)%mod;
printf("%lld ",ans);
break;
}
m-=sb-i+1;
}
}
return 0;
}
T4
密码学??!
不会,据说好像是找某几个特殊字符,然后一一确定
不写了不写了
标签:ck,11.08,noip,int,comp,mid,ans,毒枭,dp 来源: https://www.cnblogs.com/hzoi-fengwu/p/15527104.html