最近有点神志无知
命运中很多事情真是奇妙
我必须改变自己的状态
1001 简单的模拟
#include<bits/stdc++.h>
using namespace std;
int main() {
int a,b;
while(~scanf("%d %d",&a,&b)) {
int tt = a+b;
if(tt == 0) {
printf("0
"); continue;
}
int fl = 1;
if(tt < 0) fl = 0;
vector<int> ans;
tt = abs(tt);
int cnt = 0;
// printf("%d
", tt);
while(tt) {
cnt ++;
if(cnt %3 == 0) ans.push_back(100+(tt%10));
else ans.push_back(tt%10);
tt /= 10;
}
if(!fl) printf("-");
for(int i = ans.size()-1; i >= 0; --i) {
if(ans[i] >= 100 && (i != ans.size()-1) ) printf(",");
if(ans[i] >= 100) ans[i] -= 100;
printf("%d",abs(ans[i]));
}
printf("
");
}
return 0;
}1002 同样模拟
#include<bits/stdc++.h>
using namespace std;
double has[1005];
int main() {
int k;
memset(has, 0, sizeof(has));
scanf("%d", &k);
for(int i = 1; i <= k; ++i) {
int a; double b; scanf("%d %lf",&a,&b);
has[a] += b;
}
scanf("%d", &k);
for(int i = 1; i <= k; ++i) {
int a; double b; scanf("%d %lf",&a,&b);
has[a] += b;
}
int all = 0;
for(int i = 0; i <= 1000; ++i) {
if(has[i] != 0) all ++;
}
printf("%d",all);
for(int i = 1000; i >= 0; --i) {
// double tt = ( (int)(has[i]*10) )*1.0 / 10;
if(has[i]) printf(" %d %.1f", i, has[i]);
}
printf("
");
}1003 我写的是改动的dijstra,我想想dij+dfs也可以
#include<bits/stdc++.h>
using namespace std;
#define sz(X) ((int)X.size())
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 505;
int rescue[N];
struct Edge{
int nx, to, dis;
}E[N*N*2];
int head[N], tot;
void add(int fr, int to, int dis) {
E[tot].to = to; E[tot].dis = dis; E[tot].nx = head[fr]; head[fr] = tot++;
}
struct Node{
int d, n, u;
Node(int a=0, int b=0, int c=0):d(a), n(b), u(c){}
bool operator < (const Node& T) const {
if(d != T.d) return d > T.d;
else return n < T.n;
}
};
int d[N]; int cnt[N]; int num[N];
int vis[N];
void dijstra(int st) {
priority_queue<Node> Q;
memset(d, INF, sizeof(d));
memset(cnt, 0, sizeof(cnt));
memset(num, 0, sizeof(num));
memset(vis, 0, sizeof(vis));
d[st] = 0; num[st] = rescue[st]; cnt[st] = 1;
for(int i = 0; i < N; ++i) cnt[i] = 1;
Q.push(Node(d[st], num[st], st));
while(!Q.empty()) {
int x = Q.top().u; Q.pop();
if(vis[x]) continue;
vis[x] = 1;
for(int i = head[x]; ~i; i = E[i].nx) {
int to = E[i].to; int fl = 1;
if(d[to] == d[x]+E[i].dis) cnt[to] += cnt[x];
if(d[to] > d[x]+E[i].dis) {
d[to] = d[x]+E[i].dis;
num[to] = num[x]+rescue[to];
cnt[to] = cnt[x];
}else if(d[to] == d[x]+E[i].dis && num[to] < num[x]+rescue[to]) {
num[to] = num[x]+rescue[to];
}else fl = 0;
if(fl) Q.push(Node(d[to], num[to], to));
}
}
}
int main() {
int n, m, c1, c2;
while(~scanf("%d %d %d %d",&n, &m, &c1, &c2)) {
memset(head, -1, sizeof(head)); tot = 0;
for(int i = 0; i < n; ++i) scanf("%d", &rescue[i]);
for(int i = 0; i < m; ++i) {
int a,b,c; scanf("%d %d %d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
dijstra(c1);
printf("%d %d
", cnt[c2], num[c2]);
}
return 0;
}1004 bfs即可
#include<bits/stdc++.h>
using namespace std;
const int N = 105;
struct Node{
int to, nx;
}E[N*N*2];
int head[N], tot;
void add(int fr, int to) {
E[tot].to = to; E[tot].nx = head[fr]; head[fr] = tot++;
}
int n,m;
int vis[N];
int du[N];
void bfs(int x) {
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(x);
vis[x] = 1;
while(!Q.empty()) {
vector<int> tmp; tmp.clear();
if(Q.front() != 1) printf(" ");
while(!Q.empty()) {
int po = Q.front(); Q.pop(); tmp.push_back(po);
}
int all = 0;
for(int i = 0; i < tmp.size(); ++i) {
int po = tmp[i];
if( (po == 1 && du[po] == 0) || (po != 1 && du[po] == 1) ) all ++;
for(int j = head[po]; ~j; j = E[j].nx) {
int to = E[j].to;
if(!vis[to]) {
vis[to] = 1;
Q.push(to);
}
}
}
printf("%d",all);
}
printf("
");
}
int main() {
while(~scanf("%d %d", &n, &m)) {
memset(head, -1, sizeof(head)); tot = 0;
memset(du, 0, sizeof(du));
for(int i = 1; i <= m; ++i) {
int a; scanf("%d",&a);
int tim; scanf("%d", &tim);
for(int j = 1; j <= tim; ++j) {
int b; scanf("%d", &b);
add(a, b); add(b, a);
du[a] ++; du[b] ++;
}
}
bfs(1);
}
return 0;
}1005 水题
#include<bits/stdc++.h>
using namespace std;
char s[105];
char a[][50] = {"zero","one","two","three","four","five","six","seven","eight","nine"};
int main() {
while(~scanf("%s", s)) {
int len = strlen(s);
int num = 0;
for(int i = 0; i < len; ++i) {
num += s[i]-'0';
}
vector<int> ans;
if(num == 0) {
printf("zero
"); continue;
}
while(num) {
ans.push_back(num%10);
num /= 10;
}
for(int i = ans.size()-1; i >= 0; --i) {
if(i != ans.size()-1) printf(" ");
printf("%s",a[ans[i]]);
}
printf("
");
}
return 0;
}1006 水题
#include<bits/stdc++.h>
using namespace std;
const int N = 105;
const int INF = 0x3f3f3f3f;
char s[N][20]; int tim[N];
int has[N];
int Time[100000];
int Tag[100000];
int main() {
int m;
while(~scanf("%d", &m)) {
memset(Time, 0, sizeof(Time));
memset(Tag, 0, sizeof(Tag));
memset(has, 0, sizeof(has));
int a1 = 0; int t1 = INF;
int a2 = 0; int t2 = -1;
for(int i = 0; i < m; ++i) {
scanf("%s", s[i]);
int a,b,c, d,e,f;
scanf("%d:%d:%d %d:%d:%d",&a,&b,&c,&d,&e,&f);
// printf("%d %d %d %d %d %d
", a, b, c, d, e, f);
int tim = a*60*60 + b*60 + c;
if(tim < t1) {
t1 = tim; a1 = i;
}
tim = d*60*60 + e*60 + f;
Time[tim] --; Tag[tim] = i+1;
if(tim > t2) {
t2 = tim; a2 = i;
}
}
printf("%s %s
", s[a1], s[a2]);
}
return 0;
}1007 大体是个dp，需要稍微改改
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+5;
const int INF = 0x3f3f3f3f;
int a[N];
int dp[N]; int st[N];
int main() {
int k;
while(~scanf("%d", &k)) {
int fl = 0;
for(int i = 0; i < k; ++i) {
scanf("%d",&a[i]);
if(a[i] >= 0) fl = 1;
}
if(fl == 0) {
printf("0 %d %d
",a[0],a[k-1]);
continue;
}
for(int i = 0; i < k; ++i) {
int pre = i?dp[i-1]:0;
if(pre <= 0) {
dp[i] = a[i]; st[i] = i;
}else {
dp[i] = pre+a[i]; st[i] = st[i-1];
}
}
int t1,t2,t3; t1 = -1;
for(int i = 0; i < k; ++i) {
if(t1 < dp[i]) {
t1 = dp[i]; t2 = st[i]; t3 = i;
}else if(t1 == dp[i] && t2 > st[i]) {
t2 = st[i]; t3 = i;
}else if(t1 == dp[i] && t2 == st[i] && t3 > i) {
t3 = i;
}
}
printf("%d %d %d
", t1, a[t2], a[t3]);
}
return 0;
}1008 水题
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+5;
const int INF = 0x3f3f3f3f;
int main() {
int n;
while(~scanf("%d", &n)) {
int pre = 0;
int ans = 0;
for(int i = 0; i < n; ++i) {
int a; scanf("%d", &a);
if(a == pre) ans += 5;
else if(a < pre) {
ans += (pre-a)*4 + 5;
}else {
ans += (a-pre)*6 + 5;
}
pre = a;
}
printf("%d
", ans);
}
return 0;
}1009
#include<bits/stdc++.h>
using namespace std;
int n1[15]; double b1[15];
int n2[15]; double b2[15];
double mp[2005];
int main() {
int k;
scanf("%d", &k);
for(int i = 1; i <= k; ++i) {
scanf("%d %lf",&n1[i],&b1[i]);
}
scanf("%d",&k);
for(int i = 1; i <= k; ++i) {
scanf("%d %lf",&n2[i],&b2[i]);
}
for(int i = 1; i <= k; ++i) {
for(int j = 1; j <= k; ++j) {
mp[n1[i]+n2[j]] += b1[i]*b2[j];
}
}
int all = 0;
for(int i = 2000; i >= 0; --i) {
if(mp[i] != 0) all++;
}
printf("%d", all);
for(int i = 2000; i >= 0; --i) {
if(mp[i] != 0) printf(" %d %.1f", i, mp[i]);
}
printf("
");
}1010 二分好题，一开始还以为会有大数，改了半天
#include<bits/stdc++.h>
using namespace std;
const int N = 105;
typedef unsigned long long ll;
const int MOD = 1e9+7;
int C(char c) {
if(c <= '9') return c-'0';
else return c-'a'+10;
}
int main() {
int tag, radix;
char s1[15]; char s2[15];
while(~scanf("%s %s %d %d", s1, s2, &tag, &radix)) {
if(tag == 2) {
for(int i = 0; i < 15; ++i) swap(s1[i], s2[i]);
}
int l1 = strlen(s1); int l2 = strlen(s2);
ll ans = 0;
for(int i = 0; i < l1; ++i) {
ans = ans*radix + C(s1[i]);
if(ans > 1e18) while(1);
}
int mincap = 0;
for(int i = 0; i < l2; ++i) {
mincap = max(mincap, C(s2[i])+1);
}
ll l = mincap; ll r = ans;
while(l <= r) {
ll mid = (l+r) >> 1;
ll tmp = 0;
for(int j = 0; j < l2; ++j) {
tmp = tmp*mid + C(s2[j]);
if(tmp > ans) {
break;
}
}
if(tmp >= ans) r = mid-1;
else l = mid+1;
}
ll tmp = 0;
for(int j = 0; j < l2; ++j) {
tmp = tmp*l + C(s2[j]);
}
if(ans != tmp) printf("Impossible
");
else printf("%lld
",l);
}
return 0;
}