Codeforces Round #598 (Div. 3)
作者：互联网
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 int main() {
5 int q; scanf("%d",&q);
6 while (q--) {
7 ll a, b, n, S;
8 scanf("%lld%lld%lld%lld",&a,&b,&n,&S);
9 if (a*n + b < S) puts("NO");
10 else if (a >= S/n) {
11 if (S - S/n*n > b) puts("NO");
12 else puts("YES");
13 }
14 else if (a < S/n) {
15 if (S - a*n > b) puts("NO");
16 else puts("YES");
17 }
18 else puts("YES");
19 }
20 return 0;
21 }
A. Payment Without Change
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 105;
4 int a[maxn], n, q;
5 bool vis[maxn];
6 int find(int num) {
7 for (int i = 1; i <= n; ++i) {
8 if (a[i] == num) return i;
9 }
10 }
11 int main() {
12 scanf("%d",&q);
13 while (q--) {
14 memset(vis,false,sizeof(vis));
15 scanf("%d",&n);
16 for (int i = 1; i <= n; ++i) {
17 scanf("%d",&a[i]);
18 }
19 for (int i = 1; i <= n; ++i) {
20 int pos = find(i);
21 for (int j = pos; j > i; --j) {
22 if (vis[j-1] == true) break;
23 if (a[j] > a[j-1]) break;
24 vis[j-1] = true;
25 swap(a[j-1],a[j]);
26 }
27 }
28 printf("%d",a[1]);
29 for (int i = 2; i <= n; ++i) {
30 printf(" %d",a[i]);
31 }
32 putchar('\n');
33 }
34 return 0;
35 }
B. Minimize the Permutation
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 10005;
4 int a[maxn], c[maxn];
5 int n, m, d;
6 int main() {
7 scanf("%d%d%d",&n,&m,&d);
8 for (int i = 1; i <= m; ++i) {
9 scanf("%d",&c[i]);
10 }
11 int pos = 0;
12 for (int i = 1; i <= m; ++i) {
13 pos = pos+d;
14 /*
15 for (int j = 0; j < c[i]; ++j) {
16 a[pos+j] = i;
17 }*/
18 pos = pos+c[i]-1;
19 }
20 if (pos + d >= n+1) {
21 puts("YES");
22 int extra = pos+d-n-1;
23 int pos = 0;
24 for (int i = 1; i <= m; ++i) {
25 if (extra == 0) pos = pos+d;
26 else {
27 if (extra >= d) pos += 1, extra -= d-1;
28 else pos += d-extra, extra = 0;
29 }
30 for (int j = 0; j < c[i]; ++j) {
31 a[pos+j] = i;
32 }
33 pos = pos+c[i]-1;
34 }
35 printf("%d",a[1]);
36 for (int i = 2; i <= n; ++i) {
37 printf(" %d",a[i]);
38 }
39 printf("\n");
40 }
41 else puts("NO");
42 return 0;
43 }
C. Platforms Jumping
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn = 1e6+5;
5 char s[maxn];
6 int n, q;
7 ll k;
8 int main() {
9 scanf("%d",&q);
10 while (q--) {
11 scanf("%d%lld",&n,&k);
12 scanf("%s",s+1);
13
14 int pos = 1;
15 for (int i = 1; i <= n && k > 0; ++i) {
16 if (s[i] == '1') continue;
17 else {
18 if (i == pos) {
19 pos++;
20 continue;
21 }
22 if (k >= i-pos) {
23 s[pos] = '0'; s[i] = '1';
24 k -= i-pos;
25 pos++;
26 }
27 else {
28 s[pos+i-pos-k] = '0'; s[i] = '1';
29 break;
30 }
31 }
32 //printf("%s\n",s+1);
33 }
34 printf("%s\n",s+1);
35 }
36 return 0;
37 }
D. Binary String Minimizing
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 2e5+5;
4 const int inf = 0x3f3f3f3f;
5 pair<int,int> a[maxn];
6 int dp[maxn], p[maxn], t[maxn];
7 int main() {
8 int n; scanf("%d",&n);
9 for (int i = 0; i < n; ++i) {
10 scanf("%d",&a[i].first);
11 a[i].second = i;
12 }
13 sort(a,a+n);
14 fill(dp+1,dp+1+n,inf);
15 fill(p+1,p+1+n,-1);
16 for (int i = 0; i < n; ++i) {
17 for (int j = 3; j <= 5 && i+j <= n; ++j) {
18 int x = a[i+j-1].first - a[i].first;
19 if (dp[i+j] > dp[i]+x) {
20 p[i+j] = i;
21 dp[i+j] = dp[i] + x;
22 }
23 }
24 }
25 int cnt = 1, pos = n;
26 while (pos) {
27 for (int i = pos-1; i >= p[pos]; --i) {
28 t[a[i].second] = cnt;
29 }
30 cnt++;
31 pos = p[pos];
32 }
33 printf("%d %d\n",dp[n],cnt-1);
34 printf("%d",t[0]);
35 for (int i = 1; i < n; ++i) {
36 printf(" %d",t[i]);
37 }
38 printf("\n");
39 return 0;
40 }
E. Yet Another Division Into Teams
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int maxn = 1e6+5;
5 char s[maxn], t[maxn];
6 int mps[300], mpt[300];
7 int C[10005];
8 int lowbit(int x) {
9 return x&(-x);
10 }
11 int sum(int x) {
12 int ret = 0;
13 while (x > 0) {
14 ret += C[x]; x -= lowbit(x);
15 }
16 return ret;
17 }
18 void add(int x, int d) {
19 while (x <= 1000) {
20 C[x] += d; x += lowbit(x);
21 }
22 }
23 int main() {
24 int q; scanf("%d",&q);
25 while (q--) {
26 int n; scanf("%d",&n);
27 scanf("%s%s",s+1,t+1);
28 memset(mps,0,sizeof(mps));
29 memset(mpt,0,sizeof(mpt));
30 for (int i = 1; i <= n; ++i) {
31 mps[s[i]]++, mpt[t[i]]++;
32 }
33 bool flag = true;
34 for (char i = 'a'; i <= 'z'; ++i) {
35 if (mps[i] != mpt[i]) flag = false;
36 }
37 if (flag == false) {
38 puts("NO");
39 continue;
40 }
41
42 memset(C,0,sizeof(C));
43 ll ans1 = 0, ans2 = 0;
44 for (int i = 1; i <= n; ++i) {
45 ans1 += i-1 - sum(s[i]);
46 add(s[i],1);
47 }
48 memset(C,0,sizeof(C));
49 for (int i = 1; i <= n; ++i) {
50 ans2 += i-1 - sum(t[i]);
51 add(t[i],1);
52 }
53 //cout << ans1 << " " << ans2 << endl;
54 if (ans1 % 2 == 1 && ans2 % 2 == 0) flag = false;
55 else if (ans1 % 2 == 0 && ans2 % 2 == 1) flag = false;
56
57 for (char i = 'a'; i <= 'z'; ++i) {
58 if (mps[i] > 1) flag = true;
59 }
60
61 if (flag == true) puts("YES");
62 else puts("NO");
63 }
64 return 0;
65 }
F. Equalizing Two Strings
标签：puts,int,598,Codeforces,pos,maxn,printf,Div,else 来源： https://www.cnblogs.com/wstong/p/11890611.html