感觉像是个人赛，不过要是个人赛的话他们真的是太牛了。。。完全被虐爆了有没有啊。
总体来说题目不是很难，下面就粘一下代码吧。
队友看的A，看完就发现了水题本质，但是敲的不是很顺利。。。是因为太久没做比赛么。。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define inf 0x3f3f3f3f
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 100086;
const int INF = 0x3f3f3f3f;
int str[maxn];
int main()
{
int n,i,j;
int m,k;
str[1]=0;
for(i=2; i<100; i++)
{
str[i]=str[i-1]+i*i;
}
str[100]=inf;
while(scanf("%d", &n) != EOF)
{
int tmp = 0;
for(i=0; i<n; i++)
{
scanf("%d%d",&m,&k);
tmp+=m*k;
}
int pos =0;
for(i=1; i<=100; i++)
{
if(tmp>=str[i])pos = i;
}
printf("Team Liserious‘ rank is %d\n",pos);
}
return 0;
}
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
int main()
{
LL b, a; int T;LL n;
scanf("%d", &T);
while(T --)
{
cin >> a >> b >> n;
LL x, y;
y = (b - a) / 2; x = a - y;
n ++;
double ans;
if((n-1)/2 > (n-2)/2) y *= 3;
ans = log10(x+y+0.0) + log10(3+0.0)*((n - 2) / 2);
printf("%.7f\n", ans);
}
} D题好像就是求一下向量夹角就行了。。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double pi=3.14159265358;
const double eps=1e-8;
int dblcmp(double x)
{
if(x<-eps) return -1;
if(x>eps) return 1;
return 0;
}
struct point
{
double x,y,k;
point(double a=0,double b=0,double c=0)
{
x=a,y=b,k=c;
}
point operator-(const point &a)
{
return point(x-a.x,y-a.y,k);
}
void input()
{
scanf("%lf%lf%lf",&x,&y,&k);
}
bool friend operator<(const point &a,const point &b)
{
return a.k<b.k;
}
} p[1010],st,en,s1,s2;
double dot(point a, point b)
{
return a.x * b.x + a.y * b.y;
}
double cross(point a, point b)
{
return a.x * b.y - a.y * b.x;
}
double angle(point a, point b)
{
double res = acos(dot(a, b)/((sqrt(a.x * a.x + a.y * a.y)*sqrt(b.x * b.x + b.y * b.y))));
if (dblcmp(cross(a, b)- 0) < 0) return 2 * pi - res;
return res;
}
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0; i<n; i++) p[i].input();
sort(p,p+n);
scanf("%lf%lf%lf%lf",&st.x,&st.y,&en.x,&en.y);
s1=p[0]-p[1];
s2=st-en;
double ans=angle(s1,s2)/pi*180.00;
if(ans>180.00) ans=360.00-ans;
printf("%.3f\n",ans);
}
}
H竟然11分钟就有人A掉了，太厉害了。。。推了下，既然是汉诺塔，就肯定是有递推关系了。t[a+1] = 1/3*t[a]+2/3*(t[a] + 1 + f[a]);f[a]表示a个盘子的正常汉诺塔问题，即本来是在一个位置A，转移到C的步数。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define MP make_pair
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 55;
const LL INF = 1ll << 60;
int main()
{
int T, n;
scanf("%d", &T);
while(T --)
{
scanf("%d", &n);
double ans = 2.0/3.0;
for(int i = 2; i <= n; i ++)
{
ans = 1.0/3.0*ans + 2.0/3.0*(ans + (1ll << (i - 1)));
}
printf("%.2f\n", ans);
}
} #include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define MP make_pair
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 100100;
const int INF = 0x3f3f3f3f;
int main()
{
int T, n, m, k;
scanf("%d", &T);
while(T --)
{
scanf("%d%d%d", &n, &m, &k);
bool flag = false;
int r = 1, c = m;
while(r <= n && c >= 1)
{
printf("%d %d\n", r, c);fflush(stdout);
int tmp; scanf("%d", &tmp);
if(tmp == k)
{
flag = true;
break;
}
if(tmp < k) r ++;
else c --;
}
if(flag) puts("YES"),fflush(stdout);
else puts("NO"),fflush(stdout);
}
}
B)((AT).+1(BT).+1)
= (A*((AT).+1))(B*((BT).+1)),这样就妥妥的能AC了。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define MP make_pair
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 55;
LL ans[3000][3000];
struct Mr
{
LL m[maxn][maxn];
int r, c;
void pt()
{
for(int i = 0; i < r; i ++)
for(int j = 0; j < c; j ++)
printf("%lld%c", m[i][j], j == c - 1? ‘\n‘ : ‘ ‘);
}
};
Mr Tadd1(Mr a)
{
Mr ret;
ret.r = a.c; ret.c = a.r;
for(int i = 0; i < a.r; i ++)
for(int j = 0; j < a.c; j ++)
{
ret.m[j][i] = a.m[i][j] + 1;
}
return ret;
}
Mr Mul(Mr a, Mr b)
{
Mr ret; ret.r = a.r; ret.c = b.c;
for(int i = 0; i < a.r; i ++)
{
for(int j = 0; j < b.c; j ++)
{
ret.m[i][j] = 0;
for(int k = 0; k < a.c; k ++)
{
ret.m[i][j] += a.m[i][k] * b.m[k][j];
}
}
}
return ret;
}
void OR(Mr a, Mr b)
{
for(int i = 0; i < a.r; i ++)
{
for(int j = 0; j < a.c; j ++)
{
for(int k = 0; k < b.r; k ++)
{
for(int s = 0; s < b.c; s ++)
{
ans[i*b.r+k][j*b.c+s] = a.m[i][j] * b.m[k][s];
}
}
}
}
}
Mr a, at, b, bt;
int main()
{
int T, n, m, p, q;
scanf("%d", &T);
while(T --)
{
scanf("%d%d%d%d", &n, &m, &p, &q);
a.r = n, a.c = m;
b.r = p, b.c = q;
for(int i = 0; i < n; i ++)
{
for(int j = 0; j < m; j ++)
scanf("%lld", &a.m[i][j]);
}
for(int i = 0; i < p; i ++)
{
for(int j = 0; j < q; j ++)
scanf("%lld", &b.m[i][j]);
}
at = Tadd1(a); bt = Tadd1(b);
OR(Mul(a, at), Mul(b, bt));
for(int i = 0; i < n*p; i ++)
{
for(int j = 0; j < n*p; j ++)
{
printf("%lld%c", ans[i][j], j == n*p - 1 ? ‘\n‘ : ‘ ‘);
}
}
return 0;
}
F题dp一枚，每次枚举出每组的情况，然后结合前面能组成的情况，dp就可以了。dp[i]表示sum个位置哪个位置被占用了。时间复杂度不明。。。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define MP make_pair
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 25;
LL dp[(1<<20)+100];
vector<int> a[maxn], now, nxt;
int sum, n, b[7];
void dfs(int idx)
{
if(idx >= n) return;
int sz = a[idx].size(), ca = 0;
for(int j = 0; j < sz; j ++)
b[j] = j, ca += a[idx][j];
b[sz] = 0;
nxt.clear();
for(int i = 0; i < sum; i ++)
{
do
{
int fu = 0, ta = i;
if(i + ca - a[idx][b[0]] >= sum) continue;
for(int j = 0; j < sz; j ++)
{
fu |= (1 << (ta));
ta += a[idx][b[j + 1]];
}
for(int j = 0; j < now.size(); j ++)
if(!(fu&now[j]))
{
if(!dp[fu|now[j]]) nxt.push_back(fu|now[j]);
dp[fu|now[j]] += dp[now[j]];
}
} while(next_permutation(b, b + sz));
}
now.clear();
for(int i = 0; i < nxt.size(); i ++) now.push_back(nxt[i]);
dfs(idx + 1);
}
int main()
{
while(scanf("%d", &n) != EOF)
{
sum = 0;
for(int i = 0; i < n; i ++)
{
int m, tmp;a[i].clear();
scanf("%d", &m);
sum += m;
while(m --)
{
scanf("%d", &tmp);
a[i].push_back(tmp);
}
}
now.clear();
CLR(dp, 0);
now.push_back(0); dp[0] = 1;
dfs(0);
printf("%lld\n", dp[(1<<sum)-1]);
}
}#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
double Dis(double x1, double y1, double x2, double y2)
{
double dx=x1-x2;
double dy=y1-y2;
return sqrt(dx*dx+dy*dy);
}
void Get_KL(double L, double x, double y, int &k, int &l, double &cd)
{
k=floor((2.0*x)/(3.0*L));
l=floor((2.0*y)/(sqrt(3.0)*L));
double d1, d2, x1, y1, x2, y2;
if ((k+l)&1)
{
x1=k*L*1.5;
y1=(l+1.0)*L*sqrt(3.0)*0.5;
x2=(k+1.0)*L*1.5;
y2=l*L*sqrt(3.0)*0.5;
d1=Dis(x1,y1, x,y);
d2=Dis(x2,y2, x,y);
if (d1>d2)
{
k++;
cd=d2;
}
else
{
l++;
cd=d1;
}
}
else
{
x1=k*L*1.5;
y1=l*L*sqrt(3.0)*0.5;
x2=(k+1.0)*L*1.5;
y2=(l+1.0)*L*sqrt(3.0)*0.5;
d1=Dis(x1,y1, x,y);
d2=Dis(x2,y2, x,y);
if (d1>d2)
{
k++,l++;
cd=d2;
}
else cd=d1;
}
}
int My_Abs(int x)
{
if (x<0) return -x;
return x;
}
int main (void)
{
double L, x1, y1, x2, y2, cd1, cd2;
int k1, l1, k2, l2, ans;
int t;
scanf("%d",&t);
while (t--)
{
scanf("%lf %lf %lf %lf %lf",&L,&y1,&x1,&y2,&x2);
y1=-y1,y2=-y2;
Get_KL(L, x1, y1, k1, l1, cd1);
Get_KL(L, x2, y2, k2, l2, cd2);
if (k1==k2&&l1==l2) printf("%d\n",0);
else
{
ans=0;
if (My_Abs(k1-k2) > My_Abs(l1-l2))
ans=My_Abs(k1-k2);
else ans=My_Abs(k1-k2)+(My_Abs(l1-l2)-My_Abs(k1-k2))/2;
printf("%d\n", ans);
}
}
return 0;
}
北京师范大学第十二届程序设计竞赛,布布扣,bubuko.com
原文：http://blog.csdn.net/ok_again/article/details/24852071