A . Calculus
Samples
Input
2
1sinx+0cosx+3x+6/sinx
0
Output
NO
YES
因为给出的级数都是发散的，所以说只需要判断系数是不是都是0就好了
只有所有都是0的时候才是可以的
int main() {
string s;
int _ = read;
while(_ --){
cin >> s;
int len = s.size(),flag = 0;
for(int i=0;i<len;i++) if(isdigit(s[i]) && s[i] != '0' && !flag) flag = 1;
if(!flag) puts("YES");
else puts("NO");
}
return 0;
}
I . License Plate Recognition
样例输入：
1
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
..........................##...........##..........#####......########..........##......########....
.......#.......#..........###..........##.........#######.....########.........###......########....
.......#.......#..........###..........##........##....##..........###.........###......##..........
.......#.......#.........####..........##...............##........###.........####......##..........
....###############......####..........##..............##........###..........####......##..........
.......#.......#.........##.#..........##..............##.......###...........####......##..........
.......#.......#.........##.##.........##..............##.......#####........#####......#######.....
.......#.......#.........##.##.........##.............##........######.......##.##......########....
.......#.......#........###.##.........##............###............##......###.##............###...
.......#########........##..##.........##...........###.............##......##..##.............##...
.......#.......#........##...#.........##...........##..............##......##..##.............##...
.......#.......#........##...##........##..........###..............##.....##...##.............##...
.......#.......#........#######........##.........###.........##....##.....#########...........##...
.......#.......#.......########........##.........##..........##....##.....#########....##.....##...
.......#########.......##....##........##........###..........###...##..........##.......##...###...
.......#.......#.......##....##........##........########......######...........##.......#######....
.......................##.....##.......##........########.......####............##........#####.....
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
....................................................................................................
样例输出：
Case #1:
5 19
24 32
40 41
50 58
63 70
76 84
89 97
当从前往后模拟wa了好几发之后，心情是十分复杂的；
当我点开题目里的链接看到->
川 这个字的时候，心情瞬间舒畅 只有这一个字不是在左右端点之内连续的
#include <iostream>
#include <stack>
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
char s[33][103];
bool check(int x){
for(int i=1;i<=30;i++) if(s[i][x] == '#') return true;
return false;
}
int main(){
int _; cin >> _;
for(int I = 1;I <= _;I ++){
printf("Case #%d:
",I);
for(int i=1;i<=30;i++) scanf("%s",s[i]+1),s[i][0] = s[i][101] = '.';
stack<PII> st;
int l = 0,r = 0,cnt = 0;
for(int i=100;i>=1;i--){
if(check(i) && !check(i+1)) {
r = i;
if(cnt == 6) break;
}
if(check(i) && !check(i-1)) {
l = i;
++ cnt;
st.push({l,r});
}
}
for(int i=1;i<=100;i++){
if(check(i) && !check(i-1)) {
l = i;
break;
}
}
st.push({l,r});
while(st.size()){
printf("%d %d
",st.top().first,st.top().second);
st.pop();
}
}
return 0;
}
B . Kanade Loves Maze Designing
Input
1
6
1 1 2 2 3
1 1 4 5 1 4
Output
495644981 518101442
495644981 518101442
397599492 896634980
612255048 326063507
495644981 518101442
397599492 896634980
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 7;
const int mod = 1e9 + 7;
const int x = 19560929;
int idx = 0,head[maxn],ct;
ll a1[maxn],a2[maxn],cnt[maxn],a[2007][2007],w[maxn];
inline ll qPow(ll a,ll b,ll k){
ll res = 0;
while(b){
if(b&1) res = (res + a) % k;
b >>= 1;
a = (a + a) % k;
}
return res;
}
struct node{
int v,nex;
}e[maxn];
void init(){
for(int i=0;i<maxn;i++) head[i] = -1;
idx = 0;
}
void dfs(int u,int fa,int rt){
for(int i=head[u];~i;i=e[i].nex){
int to = e[i].v;
if(to == fa) continue;
if(cnt[w[to]] == 0) ++ ct;
++ cnt[w[to]];
a[rt][to] = ct;
dfs(to,u,rt);
-- cnt[w[to]];
if(cnt[w[to]] == 0) -- ct;
}
}
void add(int u,int v){
e[idx].v = v;
e[idx].nex = head[u];
head[u] = idx++;
}
int main(){
a1[1] = a2[1] = 1;
for(int i=2;i<=2000;i++) a1[i] = qPow(a1[i-1],x,mod),a2[i] = qPow(a2[i-1],x,mod+2);
int _; cin >> _;
while(_ --){
init();
memset(cnt,0,sizeof cnt);
int n;
scanf("%d",&n);
for(int i=2;i<=n;i++){
int v;scanf("%d",&v);
add(i,v);
add(v,i);
}
for(int i=1;i<=n;i++) scanf("%lld",&w[i]);
for(int i=1;i<=n;i++){
ct = 1;
++ cnt[w[i]];
a[i][i] = 1;
dfs(i,0,i);
-- cnt[w[i]];
}
for(int i=1;i<=n;i++){
ll c1 = 0,c2 = 0;
for(int j=1;j<=n;j++){
c1 = (c1 + a[i][j] * a1[j] % mod) % mod;
c2 = (c2 + a[i][j] * a2[j] % (mod + 2)) % (mod + 2);
}
printf("%lld %lld
",c1,c2);
}
}
return 0;
}
H . Lawn of the Dead
输入
1
4 4 4
1 3
3 4
3 2
4 3
输出
10
给出一个 n * m的方格，其中有k个位置不能走，开始在(1,1)位置上
每次移动只能是往下或者是右移动，问能够走到的格子的数量是多少
跟lc哥学线段树 链接
#define PI (double)acos(-1.0)
#define debug(x) cout<<#x<<" :"<<x<<endl
#define IOS ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mid ((l+r)>>1)
struct SegmentTree{
struct node{
int l,r,len,sum,lazy;
}tree[maxn << 2];
void Pushup(int rt){
tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}
void Pushdown(int rt){
if(tree[rt].lazy != -1) {
int lz = tree[rt].lazy;
tree[rt].lazy = -1;
tree[rt<<1].lazy = tree[rt<<1|1].lazy = lz;
tree[rt<<1].sum = tree[rt<<1].len * lz;
tree[rt<<1|1].sum = tree[rt<<1|1].len * lz;
}
}
void Build(int rt,int l,int r){
tree[rt] = {l,r,r-l+1,0,-1};
if(l == r) return ;
Build(rt<<1,l,mid);
Build(rt<<1|1,mid+1,r);
}
void Update(int rt,int l,int r,int x){
if(tree[rt].l > r || tree[rt].r < l) return ;
if(tree[rt].l >= l && tree[rt].r <= r) {
tree[rt].lazy = x;
tree[rt].sum = tree[rt].len * x;
return ;
}
Pushdown(rt);
Update(rt<<1,l,r,x);
Update(rt<<1|1,l,r,x);
Pushup(rt);
}
int QuerySum(int rt,int l,int r){
if(tree[rt].l > r || tree[rt].r < l) return 0;
if(tree[rt].l >= l && tree[rt].r <= r) return tree[rt].sum;
Pushdown(rt);
return QuerySum(rt<<1,l,r) + QuerySum(rt<<1|1,l,r);
}
int Query(int rt,int l,int r){
if(l > r || QuerySum(rt,l,r) == 0) return -1;
if(tree[rt].l == tree[rt].r) return tree[rt].l;
Pushdown(rt);
int lval = QuerySum(rt<<1,l,r);
int rval = QuerySum(rt<<1|1,l,r);
if(lval) return Query(rt<<1,l,r);
else return Query(rt<<1|1,l,r);
}
}T[2];
vector<int> vet[maxn];
int main() {
int _ = read;
while(_ --){
int n = read,m = read,k = read;
for(int i=1;i<=m;i++) vet[i].clear(),vet[i].push_back(0);
while(k --){
int x = read,y = read;
vet[y].push_back(x);
}
ll ans = 0;
T[0].Build(1,1,n);
T[1].Build(1,1,n);
sort(vet[1].begin(),vet[1].end());
if(vet[1].size() >= 2) T[1].Update(1,1,vet[1][1]-1,1);
else T[1].Update(1,1,n,1);
ans += T[1].tree[1].sum;
for(int i=2;i<=m;i++){
int cur = i&1,pre = cur^1;
T[cur].Update(1,1,n,1);
int siz = vet[i].size();
for(int j=0;j<siz;j++){
int at = vet[i][j];
int val = T[pre].Query(1,at+1,n);
if(val == -1) T[cur].Update(1,at,n,0);
else T[cur].Update(1,at,val-1,0);
}
ans += T[cur].tree[1].sum;
}
printf("%lld
",ans);
}
return 0;
}