第十三届蓝桥杯大赛软件赛省赛 C/C++ 大学 B 组小结
提示：代码为比赛时本人提交的代码，不代表是正解！！！
试题 A: 九进制转十进制
暴力
1478
试题 B: 顺子日期
012居然不算QAQ！！！！！！！
反正也是暴力，根据队友的答案应该是
4
试题 C: 刷题统计
假设需要用x周+y天，我们可以快速求出需要用多少周，再慢慢减求出多出几天就行
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const ll N = 1e5+9;
int main(){
ll a,b,n;
scanf("%lld %lld %lld",&a,&b,&n);
ll num = n/(5*a+2*b);
n = n - num*(5*a+2*b);
num = num*7;
for(int i = 1;i <= 7;i++){
if(n <= 0) break;
if(i <= 5){
n -= a;
num++;
}else{
n -= b;
num++;
}
}
printf("%lld",num);
return 0;
}
试题 D: 修剪灌木
第i个灌木如何长最高呢，那就是爱丽丝从当前灌木往左剪回来i，和往右剪回来i，看看哪个高。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const ll N = 1e5+9;
int n;
int a[N];
int maxx[N];
int main(){
scanf("%d",&n);
for(int i = 1;i <= n;i++){
maxx[i] = max(2*(i-1),2*(n-i));
}
for(int i = 1;i <= n;i++){
printf("%d\n",maxx[i]);
}
return 0;
}
试题 E: X 进制减法
因为a是大于b的，所以想要让a-b尽可能小，就要让每一位的进制尽可能小。
pa[i]代表的是第i+1位数1转化为十进制的大小
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const ll N = 1e6+9;
const ll mod = 1e9+7;
ll pow_mod(ll a,ll b){
ll ans = 1;
ll base = a;
while(b){
if(b&1) ans = (ans * base) % mod;
base = (base * base) % mod;
b >>= 1;
}
return ans;
}
ll n;
ll a[N];
ll b[N];
ll na,nb;
ll pa[N];
int main(){
scanf("%lld",&n);
scanf("%lld",&na);
for(ll i = 1;i <= na;i++){
scanf("%lld",&a[i]);
}
reverse(a+1,a+1+na);
scanf("%lld",&nb);
for(ll i = 1;i <= nb;i++){
scanf("%lld",&b[i]);
}
reverse(b+1,b+1+nb);
ll p = max(na,nb);
ll ans = 0;
pa[0] = 1;
for(ll i = 1;i <= p;i++){
ll maxx = max(a[i],b[i]);
ll x = max(2ll,maxx + 1);
pa[i] = (pa[i-1]*x)%mod;
ans = (ans + a[i]*pa[i-1] - b[i]*pa[i-1] + mod) % mod;
}
printf("%lld",ans);
return 0;
}
试题 F: 统计子矩阵
比赛时算错复杂度，用二维前缀和暴力算的。70分
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const ll N = 5e2+9;
ll a[N][N];
ll pre[N][N];
ll work(int x1,int y1,int x2,int y2){
return pre[x2][y2] - pre[x1-1][y2] - pre[x2][y1-1] + pre[x1-1][y1-1];
}
int main(){
ll ans = 0;
ll n,m,k;
scanf("%lld %lld %lld",&n,&m,&k);
for(ll i = 1;i <= n;i++){
for(ll j = 1;j <= m;j++){
scanf("%lld",&a[i][j]);
pre[i][j] = pre[i-1][j] + pre[i][j-1] - pre[i-1][j-1] + a[i][j];
}
}
for(int i = 1;i <= n;i++){
for(int j = 1;j <= m;j++){
for(int p = i;p <= n;p++){
for(int l = j;l <= m;l++){
ll x = work(i,j,p,l);
if(x <= k) ans++;
}
}
}
}
printf("%lld",ans);
return 0;
}
试题 G: 积木画
不会！！！！！！！
试题 H: 扫雷
根据雷和雷的关系建一张有向图，注意因为r最大是10，所以最多只用在平面上找20*20个点，用map存图，比赛时没算复杂度感觉能过，现在算了一下，50000 * 20 * 20 * log(50000)有点极限，希望测评机够快。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const ll N = 1e6+9;
struct node{
ll x,y,r;
}e[N];
vector<int> g[N];
map<pair<ll,ll> ,int> mp;
ll dis(ll x1,ll y1,ll x2,ll y2){
return (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2);
}
int vis[N];
void dfs(int now){
if(vis[now]) return ;
vis[now] = 1;
for(int i = 0;i < g[now].size();i++){
int to = g[now][i];
if(!vis[to]) dfs(to);
}
}
int main(){
ll n,m,r;
scanf("%lld %lld %lld",&n,&m,&r);
for(ll i = 1;i <= n;i++){
scanf("%lld %lld %lld",&e[i].x,&e[i].y,&e[i].r);
mp[make_pair(e[i].x,e[i].y)] = i;
}
for(int k = 1;k <= n;k++){
for(int i = -e[k].r;i <= e[k].r;i++){
for(int j = -e[k].r;j <= e[k].r;j++){
ll xx = e[k].x+i,yy = e[k].y+j;
if(dis(e[k].x,e[k].y,xx,yy) <= e[k].r*e[k].r){
if(mp[make_pair(xx,yy)]){
g[k].push_back(mp[make_pair(xx,yy)]);
}
}
}
}
}
for(ll k = 1;k <= m;k++){
for(int i = -e[k].r;i <= e[k].r;i++){
for(int j = -e[k].r;j <= e[k].r;j++){
ll xx = e[k].x+i,yy = e[k].y+j;
if(dis(e[k].x,e[k].y,xx,yy) <= e[k].r*e[k].r){
if(mp[make_pair(xx,yy)]){
int now = mp[make_pair(xx,yy)];
if(!vis[now])
dfs(now);
}
}
}
}
}
ll ans = 0;
for(int i = 1;i <= n;i++){
if(vis[i]) ans++;
}
printf("%lld",ans);
return 0;
}
试题 I: 李白打酒加强版
三维dp，定义d[i] [j] [k]表示前i个去处，去了j家店，还剩余k斗酒有多少种情况。
因为最后一定要遇到花，所以d[n+m] [n] [0]一定只能由d[n+m-1] [n] [1]转移而来，所以输出d[n+m-1] [n] [1]即可
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const ll N = 2e2+9;
const ll mod = 1000000007;
ll d[N][N][N];
ll n,m;
int main(){
d[0][0][2] = 1;
scanf("%lld %lld",&n,&m);
ll ans = 0;
for(int i = 1;i <= n+m;i++){
for(int j = 0;j <= n;j++){
for(int k = 0;k <= m;k++){
if(i != n+m){
if(d[i-1][j][k]){
if(k > 0){
d[i][j][k-1] = (d[i][j][k-1] + d[i-1][j][k])%mod;
}
if(k*2 <= m){
d[i][j+1][k*2] = (d[i][j+1][k*2] + d[i-1][j][k])%mod;
}
}
}
}
}
}
printf("%lld",d[n+m-1][n][1]);
return 0;
}
试题 J: 砍竹子
写完i没时间看了QAQ