D - What a Beautiful Lake

找遞增長度遞減長度-1

# include<bits/stdc++.h> #define maxn 205 using namespace std; int n; int a[maxn]; int main(){//freopen("in.txt","r",stdin);while(scanf("%d",&n)!=EOF){if(!n) break;for(int i=0;i<n;i++) scanf("%d",&a[i]),a[i+n]=a[i];int res=0;for(int i=0;i<2*n;i++)for(int j=i;j<2*n;j++){bool flg1=1,flg2=1;for(int k=i+1;k<=j;k++){if(a[k]<=a[k-1]) flg1=false;if(a[k]>=a[k-1]) flg2=false;}if(flg1||flg2)res=max(res,j-i);}res=min(res,n-1);cout<<res<<endl;}return 0; }

E - What a Ridiculous Election

雙向寬搜。

#include <bits/stdc++.h>using namespace std; typedef long long ll;int vis[100000][4][3];struct node{int now,t2,t3,step;node(){}node(int _now,int _t2,int _t3,int _step):now(_now),t2(_t2),t3(_t3),step(_step){} };queue<node> Q;int base[10],top;inline int get1(int now,int i){memset(base,0,sizeof base);top=0;while(top<=4){base[++top]=now%10;now/=10;}swap(base[i],base[i+1]);int ret=0;for(int j=5;j>=1;j--)ret=ret*10+base[j];return ret; }inline int get2(int now,int i){memset(base,0,sizeof base);top=0;while(top<=4){base[++top]=now%10;now/=10;}base[i]=(base[i]+1)%10;int ret=0;for(int j=5;j>=1;j--)ret=ret*10+base[j];return ret; }inline int get3(int now,int i){memset(base,0,sizeof base);top=0;while(top<=4){base[++top]=now%10;now/=10;}base[i]=(base[i]*2)%10;int ret=0;for(int j=5;j>=1;j--)ret=ret*10+base[j];return ret; }void init(){memset(vis,63,sizeof vis);Q.push(node(12345,3,2,0));vis[12345][3][2]=0;while(!Q.empty()){int now=Q.front().now;int t2=Q.front().t2;int t3=Q.front().t3;int step=Q.front().step;Q.pop();//cout<<t2<<" "<<t3<<endl;//1for(int i=1;i<=4;i++){int newt=get1(now,i);if(vis[newt][t2][t3]<=step+1) continue;vis[newt][t2][t3]=step+1;Q.push(node(newt,t2,t3,step+1));}//2if(t2>0){for(int i=1;i<=5;i++){int newt=get2(now,i);if(vis[newt][t2-1][t3]<=step+1) continue;vis[newt][t2-1][t3]=step+1;Q.push(node(newt,t2-1,t3,step+1));}}//3if(t3>0){for(int i=1;i<=5;i++){int newt=get3(now,i);if(vis[newt][t2][t3-1]<=step+1) continue;vis[newt][t2][t3-1]=step+1;Q.push(node(newt,t2,t3-1,step+1));}}} }int main(){//freopen("in.txt","r",stdin);init();//cout<<get1(2345,1)<<endl;int n;char st[20];while(~scanf("%s",&st)){n=0;for(int i=0;i<(int)strlen(st);i++)n=n*10+st[i]-'0';int Min=vis[0][0][0];for(int i=0;i<=3;i++)for(int j=0;j<=2;j++)Min=min(Min,vis[n][i][j]);if(Min==vis[0][0][0]) Min=-1;printf("%d\n",Min);}return 0; }

F - What a Simple Research

模擬。

#include <bits/stdc++.h>using namespace std; struct node {char c;int num; };bool cmp(node a,node b) {if(a.num==b.num) return a.c<b.c;else return a.num>b.num; } int m,n; node a[10]; char s[200]; int main() {//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);while(scanf("%d%d",&n,&m)!=EOF) {if(n==0&&m==0) break;a[0].c='A';a[1].c='C';a[2].c='D';a[3].c='E';a[4].c='G';a[0].num=0;a[1].num=0;a[2].num=0;a[3].num=0;a[4].num=0;for(int i=1;i<=n;i++) {scanf("%s",s);//printf("%s\n",s);for(int j=0;j<m;j++) {if(s[j]=='A') a[0].num++;if(s[j]=='C') a[1].num++;if(s[j]=='D') a[2].num++;if(s[j]=='E') a[3].num++;if(s[j]=='G') a[4].num++;}}sort(a,a+5,cmp);int flg=0;for(int i=0;i<5;i++) {if(a[i].num) {if(flg) printf(" ");flg=1;printf("%c %d",a[i].c,a[i].num);}}printf("\n");}return 0; }

I - A Boring Problem

前項向后項轉移可以有低階的利用二項式定理展開，需要注意維護的是前綴k次方和，求的是和的k次方。

#include <bits/stdc++.h>using namespace std; typedef long long ll; const int mod = 1000000007; ll C[200][200];ll pow1[200],pow2[200],pp,p1,p2;int n,k;char st[50005]; ll S[50005]; ll ans[50005]; ll PM[50005][105];void init(){C[0][0]=1;for(int i=1;i<=150;i++){C[i][0]=1;for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;} }int main(){//freopen("in.txt","r",stdin);init();int T;cin>>T;while(T--){scanf("%d%d",&n,&k);scanf("%s",st+1);int len=strlen(st+1);for(int i=1;i<=len;i++)S[i]=S[i-1]+st[i]-'0',PM[i][0]=S[i];for(int i=1;i<=len;i++)for(int j=1;j<=k;j++)PM[i][j]=PM[i][j-1]*PM[i][0]%mod;for(int i=1;i<=len;i++)for(int j=0;j<=k;j++)PM[i][j]=(PM[i][j]+PM[i-1][j])%mod;p1=p2=0;for(int i=1;i<=len;i++){ans[i]=0;p1=S[i];pow1[0]=1;for(ll c=1;c<=k;c++)pow1[c]=pow1[c-1]*p1%mod;ans[i]=pow1[k]*i%mod;for(int c=1;c<=k;c++){ll ft=1;if(c&1) ft=-1;ans[i]=(ans[i]+C[k][c]*pow1[k-c]%mod*PM[i-1][c-1]%mod*ft)%mod;}if(ans[i]<0) ans[i]+=mod;}for(int i=1;i<=n;i++)printf("%lld%c",ans[i]," \n"[i==n]);}return 0; }

K - JiLi Number

大約到1e9的時候就幾乎沒有這類數了，打表找出來。

#include <bits/stdc++.h>using namespace std; typedef long long ll;const ll d[100]={0 ,1 ,199981 ,199982 ,199983 ,199984 ,199985 ,199986 ,199987 ,199988 ,199989 ,199990 ,200000 ,200001 ,1599981 ,1599982 ,1599983 ,1599984 ,1599985 ,1599986 ,1599987 ,1599988 ,1599989 ,1599990 ,2600000 ,2600001 ,13199998 ,35000000 ,35000001 ,35199981 ,35199982 ,35199983 ,35199984 ,35199985 ,35199986 ,35199987 ,35199988 ,35199989 ,35199990 ,35200000 ,35200001 ,117463825 ,500000000 ,500000001 ,500199981 ,500199982 ,500199983 ,500199984 ,500199985 ,500199986 ,500199987 ,500199988 ,500199989 ,500199990 ,500200000 ,500200001 ,501599981 ,501599982 ,501599983 ,501599984 ,501599985 ,501599986 ,501599987 ,501599988 ,501599989 ,501599990 ,502600000 ,502600001 ,513199998 ,535000000 ,535000001 ,535199981 ,535199982 ,535199983 ,535199984 ,535199985 ,535199986 ,535199987 ,535199988 ,535199989 ,535199990 ,535200000 ,535200001 ,1111111110 };int main(){//freopen("in.txt","r",stdin);char st[205];while(~scanf("%s",st)){int len = strlen(st);if(len>11){puts("83 1111111110");continue;}ll n=0;for(int i=0;i<len;i++)n=n*10+st[i]-'0';int k=83;for(;k>=1;k--){if(d[k]<=n) break;}printf("%d %lld\n",k,d[k]);}return 0; }

轉載于:https://www.cnblogs.com/foreignbill/p/7875902.html

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