//chord
//copyright by ametake
#include#include#include#includeusing namespace std;
int xa,ya,xb,yb,ra,rb;
double ans;
int main()
{
freopen("chord.in","r",stdin);
freopen("chord.out","w",stdout);
//freopen("1.txt","r",stdin);
//freopen("2.txt","w",stdout);
scanf("%d%d%d",&xa,&ya,&ra);
scanf("%d%d%d",&xb,&yb,&rb);
ans=sqrt((double)(xb-xa)*(double)(xb-xa)+(double)(yb-ya)*(double)(yb-ya));
ans*=2;
printf("%.6f\n",ans);
return 0;
}

//chord
//copyright by ametake
#include#include#include#includeusing namespace std;
int xa,ya,xb,yb,ra,rb;
double ans;
int main()
{
freopen("chord.in","r",stdin);
freopen("chord.out","w",stdout);
//freopen("1.txt","r",stdin);
//freopen("2.txt","w",stdout);
scanf("%d%d%d",&xa,&ya,&ra);
scanf("%d%d%d",&xb,&yb,&rb);
ans=sqrt((double)(xb-xa)*(double)(xb-xa)+(double)(yb-ya)*(double)(yb-ya));
int r=max(ra,rb);
if (ans<=r) ans=r;
ans*=2;
printf("%.6f\n",ans);
return 0;
}

//circle
//copyright by ametake
//思路 从基数开始一个个尝试基数的倍数 第一次快速幂 之后高精乘即可 每次检验
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>using namespace std;const int sx=600;
const int maxn=100+10;
//const int power=1;
const int base=10;
const int jishu[10]={1,1,4,4,2,1,1,4,4,2};int a[maxn],ans[maxn],c[maxn];
char cc[maxn];
int now[maxn],jj[maxn];
int k;void multi(int *aa,int *b)
{memset(c,0,sizeof(c));int len1=aa[0];int len2=b[0];c[0]=min(100,aa[0]+b[0]-1);for (int i=1;i<=len1;i++){for (int j=1;j<=min(len2+i-1,100-i+1);j++){c[i+j-1]+=aa[i]*b[j];c[i+j]+=c[i+j-1]/base;c[i+j-1]%=base;}}if (c[c[0]+1]) if (c[0]<100) c[0]++;memcpy(aa,c,sizeof(c));
}bool check()
{memcpy(ans,now,sizeof(now));multi(ans,a);bool noo=false;for (int i=1;i<=k;i++) if (ans[i]!=a[i]) noo=true;if (noo) return false;else return true;
}int main()
{//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);freopen("circle.in","r",stdin);freopen("circle.out","w",stdout);scanf("%s",cc);scanf("%d",&k);int p=0;//p是基数，a是最原始数组，now是当前数组，由a求幂而来且指数一定为基数的倍数，ans是检验时用的要再乘一遍a now[0]=k;a[0]=k;if (k>strlen(cc)){printf("-1\n");return 0;}for (int i=strlen(cc);i>=1;i--){a[++p]=cc[i-1]-'0';now[p]=cc[i-1]-'0';if (p==k) break;}p=jishu[now[1]];//从此a不动 if (p>1) multi(now,now);if (p==4) multi(now,now);memcpy(jj,now,sizeof(now));//jj是基底数组 每次乘它 if (check()){printf("%d\n",p);return 0;} else{for (int i=2;i<=sx/p;i++){multi(now,jj);if (check()){printf("%d\n",p*i);return 0;} }}printf("-1\n");return 0;
}

这道题正解至今不明白为什么是十次，只是知道每次对于某一位向前递推最后求乘积···这个大家自己去网上搜吧

//altitude
//copyright by ametake
#include#include#includeusing namespace std;
const int maxn=100+10;
int a[maxn][maxn];
int n,s,k;
inline int read()
{
int f=1,a=0;
char ch=getchar();
while (ch<'0'||ch>'9')
{
if (ch=='-') f=-1;
ch=getchar();
}
while (ch>='0'&&ch<='9')
{
a=a*10+ch-48;
ch=getchar();
}
a*=f;
return a;
}
int main()
{
//freopen("1.txt","r",stdin);
//freopen("2.txt","w",stdout);
freopen("altitude.in","r",stdin);
freopen("altitude.out","w",stdout);
n=read();
s=read();
k=read();
int x,y;
for (int i=1;i<=n;i++)
{
for (int j=1;j<=n;j++)
{
a[i][j]=read();
}
}
for (int i=1;i<=k;i++)
{
x=read();
y=read();
int mi=0x3f3f3f;
int ma=0;
for (int i=x;i<=x+s-1;i++)
{
for (int j=y;j<=y+s-1;j++)
{
ma=max(a[i][j],ma);
mi=min(a[i][j],mi);
}
}
printf("%d\n",ma-mi);
}
return 0;
}

也可以用DP做，但时间复杂度更大，因为DP球了所有的状态

//sudoku
//copyright by ametake
#include#include#include#define dfsnext(x,y) {y==9?dfs(x+1,1):dfs(x,y+1);}
using namespace std;
int a[10][10];//ËÑË÷Êý×é
bool b[10][10];//ÊÇ·ñÓÐԭʼÊý×Ö ÓÐÊÇtrue true²»ÄÜÌîÊý
bool h[10][10],l[10][10],lit[10][10];//true²»ÄÜÌî
bool ok=false;
bool can(int x,int y,int num)
{
if (h[x][num]) return false;
if (l[y][num]) return false;
if (lit[(x-1)/3*3+(y-1)/3+1][num]) return false;
h[x][num]=true;
l[y][num]=true;
lit[(x-1)/3*3+(y-1)/3+1][num]=true;
a[x][y]=num;
return true;
}
void del(int x,int y,int num)
{
h[x][num]=l[y][num]=lit[(x-1)/3*3+(y-1)/3+1][num]=false;
a[x][y]=0;
}
void print()
{
for (int i=1;i<=9;i++)
{
for (int j=1;j<=9;j++)
{
printf("%d",a[i][j]);
}
printf("\n");
}
}
void dfs(int x,int y)
{
if (ok) return;
if (x==10&&y==1)
{
ok=true;
print();
return;
}
if (b[x][y]) dfsnext(x,y);
if (!b[x][y])
{
for (int i=1;i<=9;i++)
{
if (can(x,y,i))
{
dfsnext(x,y);
if (ok) break;
del(x,y,i);
}
}
}
}
int main()
{
freopen("sudoku.in","r",stdin);
freopen("sudoku.out","w",stdout);
char s[10];
for (int i=1;i<=9;i++)
{
scanf("%s",s);
for (int j=1;j<=9;j++)
{
if (s[j-1]=='?') a[i][j]=0;
else
{
a[i][j]=s[j-1]-'0';
h[i][s[j-1]-'0']=true;
l[j][s[j-1]-'0']=true;
lit[(i-1)/3*3+(j-1)/3+1][s[j-1]-'0']=true;
b[i][j]=true;
}
}
}
dfs(1,1);
return 0;
}