Codeforces Round #289 (Div. 2, acm icpc Rules) (A, B, C, E ),

Codeforces Round #289 (Div. 2, acm icpc Rules) (A, B, C, E ),

A: Water question. You can pre-process the question and output it.

B: First find out the largest and smallest parts, so that the smallest part is 1, and then the largest part is 1, 3, 4, 5... so if MAX-MIN> k is NO, otherwise the answer will be constructed based on this

C: greedy strategy. To keep the number as small as possible, compare it with the previous number. If the sum is smaller than the previous one, find a new number first, make the sum smaller than the required number, and the number is greater than the minimum value of the previous number. The method is to keep putting 0 places at the end. Now, the only difference is the need for a bigger sum. This is greedy, from the end to 9 as much as possible.

E: A counting problem. In fact, you only need to consider the value added to each letter. For each letter, the length that can reach the leftmost and rightmost is + 1, but the largest one in the middle is: 1/I + 1/(I + 1) + 1/(I + 2 )...., I is the value of the distance between the number and the two sides. For example, if the total length is 7 and the current position is 2, then I is from 3 to 6, because there are more than 6 following, you only need to add the sum of 1/(n-I + 1) * I calculated from the following forward, so we need to pre-process the two Arrays for counting, specific Code:

A:

#include <cstdio>#include <cstring>using namespace std;int n, a[15][15];int main() {    for (int i = 1; i <= 10; i++) a[1][i] = a[i][1] = 1;    for (int i = 2; i <= 10; i++)        for (int j = 2; j <= 10; j++)            a[i][j] = a[i - 1][j] + a[i][j - 1];    while (~scanf("%d", &n)){        printf("%d\n", a[n][n]);    }    return 0;}

B:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, k, a[105];int main() {    while (~scanf("%d%d", &n, &k)) {        int Min = 105, Max = 0;        for (int i = 0; i < n; i++) {            scanf("%d", &a[i]);            Min = min(Min, a[i]);            Max = max(Max, a[i]);        }        int use = min(k, Min);        int one = Min / use;        int yu = Min % use;        if (Max - Min > k) printf("NO\n");        else {            printf("YES\n");            for (int i = 0; i < n; i++) {                printf("1");                for (int j = 2; j <= Min; j++)                    printf(" 1");                for (int j = Min + 1; j <= a[i]; j++)                    printf(" %d", j - Min);                printf("\n");            }        }    }    return 0;}

C:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 305;int n, b[N];int num[10005], nn;void find(int cha) {    for (int i = 0; i < nn; i++) {        if (cha > 0) {            num[i] += 1;            for (int j = i; j < nn; j++) {                num[j + 1] += num[j] / 10;                num[j] %= 10;            }            if (num[nn]) nn++;            return;        }        cha += num[i];        num[i] = 0;    }    num[nn++] = 1;}int main() {    while (~scanf("%d", &n)) {        nn = 1;        memset(num, 0, sizeof(num));        for (int i = 0; i < n; i++)            scanf("%d", &b[i]);        for (int i = 0; i < n; i++) {            int tmp = 0;            for (int i = 0; i < nn; i++)                tmp += num[i];            int cha = b[i] - tmp;            if (cha <= 0) find(cha);            tmp = 0;            for (int i = 0; i < nn; i++)                tmp += num[i];            cha = b[i] - tmp;            for (int i = 0; i < nn; i++) {                if (num[i] + cha <= 9) {                    num[i] += cha;                    cha = 0;                    break;                } else {                    cha = max(cha - 9 + num[i], 0);                    num[i] = 9;                }            }            if (cha) {                while (cha) {                    num[nn++] = min(cha, 9);                    cha = max(cha - 9, 0);                }            }            for (int i = nn - 1; i >= 0; i--)                printf("%d", num[i]);            printf("\n");        }    }    return 0;}

E:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 500005;char str[N];double cal1[N], cal2[N];int vis[1005];double ans;int main() {    gets(str + 1);    int n = strlen(str + 1);    for (int i = 1; i <= n; i++) {        cal1[i] = cal1[i - 1] + 1.0 / i;        cal2[i] = cal2[i - 1] + 1.0 / (n - i + 1) * i;    }    ans = 0;    memset(vis, 0, sizeof(vis));    vis['I'] = vis['E'] = vis['A'] = vis['O'] = vis['U'] = vis['Y'] = 1;    for (int i = 1; i <= n; i++) {        if (vis[str[i]]) {            int c = min(i,  n - i + 1);            ans += c * (cal1[n - c + 1] - cal1[c]) + c + cal2[c - 1];        }    }    printf("%.10lf\n", ans);    return 0;}