A.字符画
签到
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2163
#include<bits/stdc++.h>
using namespace std;
int w;
void go(){
for(int i=1;i<=w;i++)
cout<<".";
}
int main(){
cin>>w;
cout<<"ooo";
go();
cout<<"ooo";
go();
cout<<"ooo";
go();
cout<<"ooo\n";
cout<<"..o";
go();
cout<<"o.o";
go();
cout<<".o.";
go();
cout<<"o.o\n";
cout<<"ooo";
go();
cout<<"o.o";
go();
cout<<".o.";
go();
cout<<"ooo\n";
cout<<"o..";
go();
cout<<"o.o";
go();
cout<<".o.";
go();
cout<<"o.o\n";
cout<<"ooo";
go();
cout<<"ooo";
go();
cout<<"ooo";
go();
cout<<"ooo\n";
return 0;
}
B.2018
打表规律
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2164
#include<bits/stdc++.h>
#define LL long long
using namespace std;
LL dp[2005][2005];
int main(){
int n,m;
for(int i=1;i<=2000;i++){
dp[1][i]=dp[i][1]=i;
}
for(int i=2;i<=2000;i++)
for(int j=2;j<=2000;j++){
dp[i][j]=dp[i-1][j]+dp[i][j-1]+1;
dp[i][j]=dp[i][j]%1000000007;
}
while(~scanf("%d %d",&n,&m)){
printf("%lld\n",dp[n][m]*dp[n][m]%1000000007);
}
return 0;
}
C.时间旅行
读题签到
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2165
#include<bits/stdc++.h>
#define LL long long
using namespace std;
int main(){
int n,m;
while(~scanf("%d %d",&n,&m)){
if(n>m) cout<<n<<"\n";
else cout<<m+1<<"\n";
}
return 0;
}
D.卖萌表情包
贪心，找到表情优先级即可
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2166
#include<bits/stdc++.h>
#define LL long long
using namespace std;
char s[1005][1005];
bool vis[1005][1005];
void go(int i,int j){
vis[i][j]=1;
}
int main(){
int n,m;
while(~scanf("%d %d",&n,&m)){
int ans=0;
memset(vis,0,sizeof(vis));
memset(s,0,sizeof(s));
for(int i=1;i<=n;i++)
scanf("%s",s[i]+1);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(vis[i][j]) continue;
if(s[i][j]==‘^‘){
if(!vis[i+1][j-1]&&!vis[i+1][j+1]&&s[i+1][j-1]==‘v‘&&s[i+1][j+1]==‘v‘){
go(i+1,j-1);
go(i+1,j+1);
go(i,j);
ans++;
}else if(!vis[i][j+2]&&!vis[i+1][j+1]&&s[i][j+2]==‘^‘&&s[i+1][j+1]==‘v‘){
ans++;
go(i,j+2);
go(i+1,j+1);
go(i,j);
}
}
else if(s[i][j]==‘<‘){
if(!vis[i-1][j+1]&&!vis[i+1][j+1]&&s[i+1][j+1]==‘>‘&&s[i-1][j+1]==‘>‘){
ans++;
go(i+1,j+1);
go(i-1,j+1);
go(i,j);
}
else if(!vis[i+1][j+1]&&!vis[i+2][j]&&s[i+1][j+1]==‘>‘&&s[i+2][j]==‘<‘){
ans++;
go(i+2,j);
go(i+1,j+1);
go(i,j);
}
}
}
}
cout<<ans<<"\n";
}
return 0;
}
J.买一送一
因为题目是一棵树，u点的贡献 == u(fa）的贡献 + 商品的总数 -（第一次出现此商品前的商品个数）
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2172
#include<bits/stdc++.h>
#define LL long long
using namespace std;
vector<int> v[100005];
int a[100005];
LL f[100005];
int vis[100005];
int pre[100005];
void dfs(int u,int cnt){
for(int i=0;i<(int)v[u].size();i++){
int to=v[u][i];
if(!vis[a[u]]) cnt++;
vis[a[u]]++;
f[to]=f[u]+cnt-pre[a[to]];
int hv=pre[a[to]];
pre[a[to]]=cnt;
dfs(to,cnt);
if(--vis[a[u]]==0) cnt--;
pre[a[to]]=hv;
}
}
int main(){
int n;
while(~scanf("%d",&n)){
for(int i=1;i<=n;i++) v[i].clear();
memset(vis,0,sizeof(vis));
memset(f,0,sizeof(f));
memset(pre,0,sizeof(pre));
for(int i=2;i<=n;i++){
int u;
scanf("%d",&u);
v[u].push_back(i);
}
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
dfs(1,0);
for(int i=2;i<=n;i++)
printf("%lld\n",f[i]);
}
return 0;
}
K.Use FFT
我们将多项式a * 多项式b 得出如下
a0*b0 a0*b1 a0*b2 a0*b3 等于a0 * b3前缀和
a1*b0 a1*b1 a1*b2 等于a1 *b2前缀和
a2*b0 a2*b1
a3*b0
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2173
#include<bits/stdc++.h>
#define LL long long
#define mod 1000000007
using namespace std;
LL a[500005],b[500005];
LL sum[1000005];
int main(){
int n,m,l,r;
while(~scanf("%d %d %d %d",&n,&m,&l,&r)){
for(int i=1;i<=n+1;i++)
scanf("%lld",&a[i]);
for(int i=1;i<=m+1;i++){
scanf("%lld",&b[i]);
sum[i]=(sum[i-1]+b[i]+mod)%mod;
}
for(int i=m+2;i<=r+1;i++)
sum[i]=sum[i-1];
LL ans=0;
for(int i=1;i<=n+1;i++){
ans=(ans+a[i]*(sum[r+1]-sum[l]+mod)%mod+mod)%mod;
if(l>0) l--;
if(r>=0) r--;
else break;
}
printf("%lld\n",(ans+mod)%mod);
}
return 0;
}
H.千万不要用树套树
对于每个查询 答案等于 -1r左边的线段个数 + l+1右边的线段个数（这样就不会重复）
我们用两个线段树维护 一个维护左线段 一个维护右线段
http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2170
#include<bits/stdc++.h>
#define LL long long
#define mod 1000000007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn = 100005;
int sum1[maxn<<2],sum2[maxn<<2],cnt[maxn];
void update1(int l,int r,int rt,int pos){
if(l==r){
sum1[rt]+=1;
return ;
}
int m=(l+r)>>1;
if(pos<=m) update1(lson,pos);
else update1(rson,pos);
sum1[rt]=sum1[rt<<1]+sum1[rt<<1|1];
}
void update2(int l,int r,int rt,int pos){
if(l==r){
sum2[rt]+=1;
return ;
}
int m=(l+r)>>1;
if(pos<=m) update2(lson,pos);
else update2(rson,pos);
sum2[rt]=sum2[rt<<1]+sum2[rt<<1|1];
}
int query1(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R){
return sum1[rt];
}
int m=(l+r)>>1;
int ans=0;
if(L<=m) ans+=query1(lson,L,R);
if(R>m) ans+=query1(rson,L,R);
return ans;
}
int query2(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R){
return sum2[rt];
}
int m=(l+r)>>1;
int ans=0;
if(L<=m) ans+=query2(lson,L,R);
if(R>m) ans+=query2(rson,L,R);
return ans;
}
int main(){
int n,q;
while(~scanf("%d %d",&n,&q)){
int all=0;
for(int i=1;i<=n*4;i++){
if(i<=n) cnt[i]=0;
sum1[i]=sum2[i]=0;
}
while(q--){
int op,l,r;
scanf("%d",&op);
if(op==1){
scanf("%d %d",&l,&r);
if(l==r) cnt[l]++;
update1(1,n,1,l);
update2(1,n,1,r);
all++;
}else {
scanf("%d %d",&l,&r);
int ans=all;
if(n>=l+1) {
ans-=query1(1,n,1,l+1,n);
}
if(r-1>=1) {
ans-=query2(1,n,1,1,r-1);
}
if(r-l==2) ans+=cnt[l+1];
printf("%d\n",ans);
}
}
}
return 0;
}
原文地址：https://www.cnblogs.com/MengX/p/10645119.html
时间： 2024-11-23 04:11:33