KRN

M:

#include <string.h>
#include <stdio.h>
int main()
{
    char a[90], b[90];
    int i, n, j = 0;
    gets(a);
    n = strlen(a);
    for (i = n - 1; i >= 0; i--)
    {
        b[j++] = a[i];
    }
    b[j] = '\0';
    puts(b);
    return 0;
}

 

N:

#include<stdio.h>

int main()
{
    int i,n,j;
    int a[15][15];
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&a[i][j]);
        int sum1=0,sum2=0;
        for(i=1;i<=n;i++){
            sum1+=a[i][i];
            sum2+=a[i][n-i+1];
        }
        printf("%d %d\n",sum1,sum2);
    }
    return 0;
}

 

O:

#include <stdio.h>
int main()
{
    int n, a[7][7], i, j;
    while (scanf("%d", &n) != EOF)
    {
        int flag = 1;
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                scanf("%d", &a[i][j]);
        for (i = 1; i < n; i++)
        {
            for (j = 0; j < i; j++)
            {
                if (a[i][j] != 0)flag = 0;
            }
        }
        if (flag)
            printf("YES\n");
        if (flag == 0)
            printf("NO\n");
    }
    return 0;
}

 

P:

#include <stdio.h>
int main()
{
    int n, a[7][7], i, j, x, y, t, k;
    while (scanf("%d", &n) != EOF)
    {
        for (i = 0; i < n; i++)
        {
            for (j = 0; j < n; j++)
                scanf("%d", &a[i][j]);
        }
        for (i = 0; i < n; i++)
        {
            x = i;
            y = 0;
            for (j = 1; j < n; j++)
                if (a[x][y] < a[i][j])
                    y = j;
            t = 1;
            for (k = 0; k < n; k++)
                if (a[k][y] < a[x][y])
                    t = 0;
            if (t == 1)
            {
                printf("%d %d\n", x, y);
                break;
            }
        }
        if (t == 0)
            printf("NO\n");
    }
    return 0;
}

Q:

#include <stdio.h>
#include <string.h>
int main()
{
    char a[80];
    int len, i;
    while (gets(a) != NULL) 
    {
        for (i = 0; a[i]!='\0'; i++)
        {
            if (a[i] >='A'&& a[i] <='Z')
            {
                a[i] ='A'+'Z'- a[i];
            }
        }
        puts(a);
    }
    return 0;
}

 

R:

#include <stdio.h>

int main()
{
    int i, n;
    int ans[1000];
    while (scanf("%d", &n) != EOF)
    {
        int cnt = 1;
        while (n)
        {
            int yu = n % 2;
            n /= 2;
            ans[cnt++] = yu;
            //printf("%d", yu);
        }
        for (int i = cnt - 1; i >= 1; i--)
        {
            printf("%d", ans[i]);
        }
        printf("\n");
    }
    return 0;
}

 #include <stdio.h>

#include <string.h>

 

const int N = 10000005;

 

int n, m, v[N], g[N], s[N];

 

void init() {

memset(v, 0, sizeof(v));

memset(g, 0, sizeof(g));

memset(s, 0, sizeof(s));

 

int a;

scanf("%d", &n);

for (int i = 0; i < n; i++) {

scanf("%d", &a);

g[a]++;

}

 

for (int i = 2; i < N; i++) {

if (v[i]) continue;

for (int j = i; j < N; j += i) {

if (g[j]) s[i] += g[j];

v[j] = 1;

}

}

for (int i = 1; i < N; i++) s[i] += s[i-1];

}

 

void solve() {

int a, b;

scanf("%d", &m);

for (int i = 0; i < m; i++) {

scanf("%d%d", &a, &b);

if (a >= N) a = N;

if (b >= N) b = N - 1;

printf("%d\n", s[b] - s[a-1]);

}

}

 

int main() {

init();

solve();

return 0;

}

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <algorithm>

 

using namespace std;

 

const double pi = acos(-1.0);

const int N = 1 << 20;

 

int n;

double r, l, dp[N];

 

struct point {

double x, y, d;

void scan() {

scanf("%lf%lf%lf", &x, &y, &d);

d = d * pi / 180; x -= l; y = fabs(y);

}

}p[N];

 

void init() {

memset(dp, 0, sizeof(dp));

scanf("%d%lf%lf", &n, &l, &r); r -= l;

for (int i = 0; i < n; i++) p[i].scan();

}

 

double f(int id, double x0) {

double tmp = atan((r - p[id].x) / p[id].y);

double du = min(tmp, atan((x0 - p[id].x) / p[id].y) + p[id].d);

return p[id].x + p[id].y * tan(du);

}

 

double solve() {

for (int i = 0; i < (1<<n); i++) {

for (int j = 0; j < n; j++) {

if (i & (1<<j)) continue;

dp[i | (1<<j)] = max(dp[i | (1<<j)], f(j, dp[i]));

}

}

return dp[(1<<n) - 1];

}

 

int main() {

init();

printf("%.9lf\n", solve());

return 0;

}