前言

開始做四面八方扣來的題

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成績

$Rank$$Person$$Score$$A$$B$$C$

$1$	$2017zyc$	$160$	$70$	$30$	$60$
$2$	$2017wyc$	$140$	$80$	$10$	$50$
$3$	$2017lrz$	$140$	$40$	$100$	$0$
$4$	$2017xjq$	$100$	$70$	$30$	$0$
$5$	$2017xxy$	$100$	$70$	$30$	$0$
$6$	$2015gjh$	$100$	$0$	$0$	$100$
$7$	$2017hjq$	$70$	$40$	$30$	$0$
$8$	$2017lw$	$70$	$40$	$30$	$0$
$9$	$2015yjy$	$70$	$40$	$30$	$0$
$10$	$2015zyf$	$70$	$40$	$0$	$30$

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正題

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$T1:nssl1194?$春思【逆元,等比數列,約數】

更之前一道題一樣的:
https://blog.csdn.net/Mr_wuyongcong/article/details/82502158

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$T2:nssl1195?$健美貓【$???$】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/83212915

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$T3:nssl1196?$摘果子【樹形依賴背包$,dp$】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/83210493

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$someofcode$

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T1 80分code

#include<cstdio> #include<algorithm> #define mod 9901 #define ll long long using namespace std; ll pr[50],c[50],a,b,cnt,ans; void prime(ll a) {for(ll i=2;i*i<=a;i++){if(!(a%i)){pr[++cnt]=i;while(!(a%i)) c[cnt]++,a/=i;}}if(a!=1) pr[++cnt]=a,c[cnt]=1; } ll power(ll a,ll b) {ll ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans; } int main() {scanf("%lld%lld",&a,&b);prime(a);ans=1;for(ll i=1;i<=cnt;i++){ll inx=power(pr[i]-1,mod-2);if(!inx){ans=ans*((c[i]*b+1)%mod)%mod;continue;}ll k=(power(pr[i],c[i]*b+1)+mod-1)%mod*inx%mod;ans=ans*k%mod;}printf("%lld",ans); }

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T2 10分code

#include<cstdio> #include<algorithm> #define N 2000010 using namespace std; int n,s[N*2],ans,sum; int main() {scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&s[i]),s[i+n]=s[i];ans=1;int i=2;while(i<=2*n){if(s[i]<s[ans]){ans=i;i++;continue;}if(s[i]>s[ans]){i++;continue;}int k=1;while(i+k<=2*n&&s[i+k]==s[ans+k]) k++;if(i+k>2*n) break;if(s[i]>s[i+k])ans=i+k;else if(s[ans+k]>s[i+k]) ans=i;i+=k+1;}for(i=ans;i<ans+n;i++)sum+=abs(s[i]-(i-ans+1));printf("%d",sum); }

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T3 50分code

#include<cstdio> #include<algorithm> #define N 2010 using namespace std; struct node{int to,next; }a[N]; int n,m,v[N],p[N],ls[N],f[N][N],tot,x,y; void addl(int x,int y) {a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot; } void dp(int x,int fa) {for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(y==fa) continue;dp(y,x);for(int j=m;j>=p[y];j--){for(int k=p[y];k<=j;k++)f[x][j]=max(f[x][j],f[x][j-k]+f[y][k]);}}for(int j=m;j>=p[x];j--)f[x][j]=f[x][j-p[x]]+v[x]; } int main() {scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d%d",&v[i],&p[i]);for(int i=1;i<n;i++){scanf("%d%d",&x,&y);addl(x,y);addl(y,x);}dp(1,0);printf("%d",f[1][m]); }

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尾聲

沒了

總結

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