2022有道小图灵信息学水平测试（总）A-X题题解
比赛地址1
比赛地址2
比赛地址3
比赛地址4
A
循环找最小值
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
double f, mx;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
mx=1e9;
for(i=1; i<=8; ++i)
scanf("%lf", &f), mx=min(mx, f);
printf("%.2lf", mx);
return 0;
}
B
循环一遍，遇到数字就输出
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1000010
int n, m, i, j, k, T;
char s[N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
scanf("%s", s+1);
for(i=1; s[i]; ++i)
if(s[i]>='0' && s[i]<='9') printf("%c", s[i]), ++k;
if(!k) printf("0");
return 0;
}
C
每次加上弹高看看高不高过篮高
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
double f;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
for(i=1; i<=n; ++i)
{
scanf("%lf", &f);
if(f+30.7>=305) ++k;
}
printf("%d", k);
return 0;
}
D
不断循环题目所说过程，然后输出，到1就停止
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
while(n!=1)
{
if(n&1) n=n*3+1;
else n/=2;
printf("%lld ", n);
}
return 0;
}
E
统计行为1或 \(n\)，列为1或 \(n\) 的和即可
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read(); m=read();
for(i=1; i<=n; ++i)
for(j=1; j<=m; ++j)
{
k=read();
if(i==1 || j==1 || i==n || j==m) T+=k;
}
printf("%lld", T);
return 0;
}
F
两个符合条件的都输出即可
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
double x, y;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
for(i=1; i<=n; ++i)
{
scanf("%lf%lf", &x, &y);
if(x<89.9 || x>150.7) continue;
if(y<49.9 || y>90.34) continue;
++k;
}
printf("%lld", k);
return 0;
}
G
输出所有数的和
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
while(n=read()) m+=n;
printf("%d", m);
return 0;
}
H
按题意所说加个判断输出即可，每个位置要么空格要么星号
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
for(i=n; i>=1; --i, printf("\n"))
for(j=n; j>=1; --j)
printf("%c", j<=i ? '*' : ' ');
return 0;
}
I
从大面值到小面值不断除模即可
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
printf("%d ", n/100), n%=100;
printf("%d ", n/50), n%=50;
printf("%d ", n/20), n%=20;
printf("%d ", n/10), n%=10;
printf("%d ", n/5), n%=5;
printf("%d", n/1);
return 0;
}
J
不算博弈的博弈，显然模4为0就输，否则赢了
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
printf(read()%4 ? "Yes" : "No");
return 0;
}
K
看看有多少个数小于等于 \(a_k\) 即可
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1010
int n, m, i, j, k, T;
int a[N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read(); k=read();
for(i=1; i<=n; ++i) a[i]=read();
for(i=1; i<=n; ++i) if(a[i]<=a[k]) ++j;
printf("%d", j);
return 0;
}
L
显然，0全丢前面，1全丢后面
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 100010
int n, m, i, j, k, T;
char a[N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
scanf("%s", a+1);
// for(i=1; <=n; ++i) a[i]=read();
sort(a+1, a+n+1);
for(i=1; i<=n; ++i) printf("%c", a[i]);
return 0;
}
M
用个map先存起来，最后判断在不在里面即可
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
//#define N
int n, m, i, j, k, T;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
T=read();
while(T--)
{
map<string, int>mp;
string s, c;
n=read();
for(i=1; i<=n; ++i)
{
cin>>s; mp[s]=1;
c+=s[0];
}
printf(mp[c] ? "yes\n" : "no\n");
}
return 0;
}
N
DP，设 \(dp(i,0/1)\) 代表前 \(i\) 枚硬币里最后一枚是反面/正面朝上的方案数
\[\Large dp(i,0)=dp(i-1,1)\\\Large dp(i,1)=dp(i-1, 0)+dp(i-1, 1)
\]
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 40
int n, m, i, j, k, T;
int dp[N][2];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
dp[1][1]=dp[1][0]=1;
for(i=2; i<=n; ++i)
{
dp[i][1]=dp[i-1][1]+dp[i-1][0];
dp[i][0]=dp[i-1][1];
}
printf("%lld", dp[n][0]+dp[n][1]);
return 0;
}
O
按题意所说排序即可
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 100010
struct node
{
int x, y, a, b, c;
string s;
}a[N];
int n, m, i, j, k, T;
bool cmp(node x, node y)
{
if(x.a==y.a) {
if(x.b==y.b) {
if(x.c==y.c) return x.s<y.s;
return x.c<y.c;
}
return x.b<y.b;
}
return x.a<y.a;
}
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
for(i=1; i<=n; ++i)
{
m=read();
for(j=1; j<=m; ++j)
{
a[++k].x=i; a[k].y=j;
cin>>a[k].s;
a[k].a=read(); a[k].b=read(); a[k].c=read();
}
}
sort(a+1, a+k+1, cmp);
for(i=1; i<=k; ++i)
cout<<a[i].s, printf(" %d %d\n", a[i].x, a[i].y);
return 0;
}
P
有点恶心
按顺序先填最左，然后斜着，然后向下，然后再斜着，填过不填就是
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 18
int n, m, i, j, k, T;
int a[N][N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
for(i=1; i<=n; ++i) a[i][1]=++k;
for(i=n, j=1; i>=1; --i, ++j)
if(!a[i][j]) a[i][j]=++k;
for(i=1; i<=n; ++i)
if(!a[i][n]) a[i][n]=++k;
for(i=n, j=n; i>=1; --i, --j)
if(!a[i][j]) a[i][j]=++k;
for(i=1; i<=n; ++i, printf("\n"))
for(j=1; j<=n; ++j)
if(a[i][j]) printf("%5lld", a[i][j]);
else printf(" ");
return 0;
}
Q
显然从后往前维护元音数量，遇到辅音就加上当前元音数量
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1000010
int n, m, i, j, k, T;
char s[N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
scanf("%s", s+1); n=strlen(s+1);
for(i=n; i>=1; --i)
{
if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u') ++k;
else m+=k;
}
printf("%lld\n", m);
return 0;
}
R
不能理解为什么答案要用复杂的二分，明明一个map就好了
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1000010
int n, m, i, j, k, T;
int a[N];
map<int, int>mp;
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read(); m=read();
for(i=1; i<=n; ++i) a[i]=read();
for(i=1; i<=m; ++i) mp[read()]=1;
for(i=1, k=0; i<=n; ++i)
if(mp[a[i]]) k=1, printf("%lld ", a[i]);
if(!k) printf("No");
return 0;
}
S
题目都告诉你是记忆化递归了，直接照着打就是了
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 25
int n, m, i, j, k, T;
int f[N][N];
int w(int x, int y)
{
if(x<=0 || y<=0) return 1;
if(x>20 || y>20) return w(20, 20);
if(f[x][y]!=-1) return f[x][y];
if(x<=y) return f[x][y]=w(x, y-1)+w(x, y-2);
else return f[x][y]=w(x-1, y)+w(x-2, y);
}
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
memset(f, -1, sizeof(f));
n=read(); m=read();
while(n!=-1 || m!=-1)
printf("%lld\n", w(n, m)), n=read(), m=read();
return 0;
}
T
**出题人公开赛考高精
知道高精就行了，套个模板直接走人
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1000
int n, m, i, j, k, T;
int a[N], b[N];
char s[N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
// k=read(); printf("%lld", k*k-k);
scanf("%s", s+1); n=strlen(s+1);
for(i=1, j=n; i<=n; ++i, --j) a[i]=s[j]-'0';
for(i=1; i<=n; ++i)
for(j=1; j<=n; ++j)
b[i+j-1]+=a[i]*a[j];
m=2*n-1;
for(i=1; i<=m; ++i)
{
if(b[i]>=10) b[i+1]+=b[i]/10, b[i]%=10;
if(b[i+1]) m=max(m, i+1);
}
while(!b[m] && m) --m;
for(i=1; i<=n; ++i) b[i]-=a[i];
for(i=1; i<=m; ++i)
if(b[i]<0) b[i]+=10, b[i+1]--;
while(!b[m] && m) --m;
// printf("%lld\n", m);
if(!m) printf("0");
else {
for(i=m; i>=1; --i) printf("%lld", b[i]);
}
return 0;
}
U
此题有点恶心
从小到大弄，不断复制即可
细节有点多
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 2050
int n, m, i, j, k, T;
char s[N][N];
int h;
// void cal(int x)
// {
// if(x==1) printf("@");
// else printf(" "), cal(x-1), printf("\n"), cal(x-1), printf(" "), cal(x-1), printf("\n");
// }
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
s[1][1]='@'; s[1][2]=' '; h=1;
for(m=1; m<n; ++m)
{
// printf("---\n");
// for(i=1; i<=h; ++i, printf("\n"))
// for(j=1; j<=(1<<n); ++j) printf("%c", s[i][j] ? s[i][j] : ' ');
// printf("--------\n");
for(i=1; i<=h; ++i)
{
for(j=1; j<=(1<<m); ++j)
s[i+h][j]=s[i+h][j+(1<<m)]=s[i][j];
}
// printf("--------\n");
for(i=1; i<=h; ++i)
for(j=(1<<(m+1)); j>=h; --j)
s[i][j]=s[i][j-h], s[i][j-h]=' ';
h<<=1;
}
// printf("--------\n");
for(i=1; i<=h; ++i, printf("\n"))
for(j=1; j<=(1<<n); ++j) printf("%c", s[i][j] ? s[i][j] : ' ');
return 0;
}
/*
@ 2^1
@
@ @ 2^2
@
@ @
@ @
@ @ @ @ 2^3
@ @
@ @ @ @
@ @ @ @
@ @ @ @ @ @ @ @
@
@ @
@ @
@ @ @ @
@ @
@ @ @ @
@ @ @ @
@ @ @ @ @ @ @ @
*/
V
因为数据很小，每个 X 都试一遍即可
注意开始结束位置相同就不用试了
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 110
int n, m, i, j, k, T;
int lx, ly, rx, ry, a[N][N];
char s[N][N];
int dx[4]={0, 0, 1, -1};
int dy[4]={1, -1, 0, 0};
int check(int x, int y, int k)
{
// printf("%d %d\n", x, y);
if(s[x][y]=='X') return 0;
if(x==rx && y==ry) return 1;
a[x][y]=k;
int b=0, i, newx, newy;
for(i=0; i<4; ++i)
{
newx=x+dx[i];
newy=y+dy[i];
if(newx<1 || newy<1 || newx>n || newy>m) continue;
if(a[newx][newy]==k) continue;
b|=check(newx, newy, k);
}
return b;
}
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read(); m=read();
for(i=1; i<=n; ++i) scanf("%s", s[i]+1);
lx=read(); ly=read(); rx=read(); ry=read();
if(lx==rx && ly==ry) return printf("0"), 0;
if(lx<1 || ly<1 || lx>n || ly>m) return printf("-1"), 0;
if(rx<1 || ry<1 || rx>n || ry>m) return printf("-1"), 0;
if(check(lx, ly, ++k)) return printf("0"), 0;
for(i=1; i<=n; ++i)
for(j=1; j<=m; ++j)
if(s[i][j]=='X')
{
s[i][j]='O';
if(check(lx, ly, ++k)) printf("%d %d\n", i, j), T=1;
s[i][j]='X';
}
if(!T) printf("-1");
return 0;
}
W
这题我一开始没看见可以停打了一坨dp上去
显然，可顺可逆，求最小，那就bfs
#include<bits/stdc++.h>
using namespace std;
//#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 2000010
struct node
{
int x, y;
bool operator <(const node &A) const
{
return x>A.x;
}
};
int n, m, i, j, k, T;
int a[N], f[N], A, B, x, y;
// priority_queue<node>q;
queue<int>q;
int s_l(int A, int B)
{
// printf("%d %d\n", A, B);
// q.push(node{-1, A});
// for(i=A; i<=B; ++i)
// {
// while(q.top().y<i) q.pop();
// f[i]=q.top().x+1;
// q.push(node{f[i], i+a[i]});
// // printf("%d ", f[i]);
// }
// while(!q.empty()) q.pop();
// printf("%d\n", f[B]);
memset(f, 0x3f, sizeof(f));
f[A]=0;
for(i=A; i<=B; ++i)
if(f[i]!=f[0] && i+a[i]<=(n<<1)) f[i+a[i]]=min(f[i+a[i]], f[i]+1);
// printf("%d\n", f[B]);
return f[B];
}
int s_r(int A, int B)
{
// printf("%d %d\n", A, B);
memset(f, 0x3f, sizeof(f));
f[A]=0;
for(i=A; i>=B; --i)
if(f[i]!=f[0] && i-a[i]>=1) f[i-a[i]]=min(f[i-a[i]], f[i]+1);
// printf("%d\n", f[B]);
return f[B];
}
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read(); A=read(); B=read();
for(i=1; i<=n; ++i) a[i]=read();
if(A==B) return printf("0");
memset(f, -1, sizeof(f));
f[A]=0; q.push(A);
while(!q.empty())
{
x=q.front(); q.pop();
y=(x+a[x]-1)%n+1;
// printf("%d ", y);
if(f[y]==-1) f[y]=f[x]+1, q.push(y);
// if(y==B) return printf("%d", f[y]), 0;
y=((x-a[x]-1)%n+n)%n+1;
if(!y) y=n;
// printf("%d ", y);
if(f[y]==-1) f[y]=f[x]+1, q.push(y);
}
printf("%d", f[B]);
// if(A<B) printf("%d", (k=min(s_l(A, B), s_r(A+n, B)))==1e9 ? -1 : k);
// if(A>B) printf("%d", (k=min(s_r(A, B), s_l(A, B+n)))==1e9 ? -1 : k);
return 0;
}
X
一看那 \(O(n^2)\) 的数据，不用动脑，盲打个dp上去就行
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define M
//#define mo
#define N 1010
int n, m, i, j, k, T;
int a[N], f[N];
signed main()
{
// freopen("tiaoshi.in", "r", stdin);
// freopen("tiaoshi.out", "w", stdout);
n=read();
for(i=1; i<=n; ++i)
{
a[i]=read(); f[i]=a[i];
for(j=1; j<i; ++j)
if(a[j]>=a[i]) f[i]=max(f[i], f[j]+a[i]);
m=max(m, f[i]);
}
printf("%lld", m);
return 0;
}
本文来自博客园，作者：zhangtingxi，转载请注明原文链接：https://www.cnblogs.com/zhangtingxi/p/16532697.html
