T1
[前缀和，差分]
求二维前缀和然后大概差分一下就好了？
【code】
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define File "wireless"
inline void file(){
freopen(File".in","r",stdin);
freopen(File".out","w",stdout);
}
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-'0'; ch = getchar();}
return x*f;
}
const int mxn = 130;
int d,n;
int s[mxn][mxn];
int a[mxn][mxn];
inline int Q(int x_1,int y_1,int x_2,int y_2){
return s[x_2][y_2] - s[x_1-1][y_2] - s[x_2][y_1-1] + s[x_1-1][y_1-1];
}
int cnt,ans;
int main(){
file();
d = read(),n = read();
for(int i = 1;i <= n; ++i){
int x,y,k;
x = read(),y = read(),k = read();
a[x][y] = k;
}
// s[0][0] = a[0][0];
for(int i = 0;i <= 128; ++i)
for(int j = 0;j <= 128; ++j)
s[i][j] = (s[i][j-1] + s[i-1][j] - s[i-1][j-1] + a[i][j]);
// ,printf("%d\n",s[i][j]);
// printf("%d\n",ans);
for(int i = 0;i <= 128; ++i){
for(int j = 0;j <= 128; ++j){
int x_1,y_1,x_2,y_2;
x_1 = (i-d >= 0) ? (i-d) : 0;
y_1 = (j-d >= 0) ? (j-d) : 0;
x_2 = (i+d <= 128) ? (i+d) : 128;
y_2 = (j+d <= 128) ? (j+d) : 128;
int t = Q(x_1,y_1,x_2,y_2);
if(t == ans) cnt++;
else if(t > ans) ans = t,cnt = 1;
}
}
printf("%d %d\n",cnt,ans);
return 0;
}
T2
[Bfs]
通过建反图判断每个点是否能到达终点。
通过枚举每个能达到终点的点，判断与之相连的边连接的点是否能出现在路径中。
最后bfs求出起点到终点的距离即可。
【code】
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define File "road"
inline void file(){
freopen(File".in","r",stdin);
freopen(File".out","w",stdout);
}
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-'0'; ch = getchar();}
return x*f;
}
const int mxn = 1e4 + 10;
int n,m;
bool v1[mxn],v2[mxn];//是否能出现在路途中。是否能到达终点。
int dis[mxn];
vector<int>E1[mxn];
vector<int>E2[mxn];
#define pb push_back
int st,ed;
queue<int> q;
int main(){
// file();
n = read(),m = read();
for(int i = 1;i <= m; ++i){
int x = read(),y = read();
E1[x].pb(y),E2[y].pb(x);
}
st = read(),ed = read();
v2[ed] = 1; q.push(ed);
while(q.size()){
int x = q.front();
q.pop();
for(int i = E2[x].size()-1; i >= 0; --i){
int y = E2[x][i];
if(!v2[y]){
q.push(y);
v2[y] = 1;
}
}
}
if(!v2[st]){
puts("-1");
return 0;
}
for(int i = 1;i <= n; ++i){
if(v2[i]){
v1[i] = 1;
for(int j = E1[i].size()-1; j >= 0; --j){
int y = E1[i][j];
if(!v2[y]){
v1[i] = 0;
break;
}
}
}
}
dis[st] = 1; q.push(st);
while(q.size()){
int x = q.front();
q.pop();
if(x == ed){
printf("%d\n",dis[ed]-1);
return 0;
}
for(int i = E1[x].size()-1; i >= 0; --i){
int y = E1[x][i];
if(v1[y] && !dis[y]){
dis[y] = dis[x] + 1;
q.push(y);
}
}
}
puts("-1");
return 0;
}
T3
[数学]
读入取模。秦九韶公式。
【code】
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define File "equation"
inline void file(){
freopen(File".in","r",stdin);
freopen(File".out","w",stdout);
}
const int mod = 998244353;
const int mxn = 110;
inline ll read(){
ll 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)%mod + (x<<3)%mod + ch-'0')%mod; ch = getchar();}
return x*f;
}
ll n,m;
ll a[mxn],b[mxn];
ll tot(0);
inline bool J(ll x){
ll s(0);
for(ll i = n; i; --i)
s = ((s + a[i])*x) % mod;
s = (s+a[0]) % mod;
return s==0;
}//秦九韶公式的判断。返回得到的s是否等于0
int main(){
file();
n = read(),m = read();
for(ll i = 0;i <= n; ++i) a[i] = read();
for(ll i = 1;i <= m; ++i){
if(J(i))
b[++tot] = i;//记录答案
}
printf("%lld\n",tot);
for(ll i = 1;i <= tot; ++i) printf("%lld\n",b[i]);
return 0;
}
我又预感因为交题交得慢我今天要ak掉rating。