CF 271 A Keyboard B Worms C Captain Marmot

A:边界不用处理，直接暴力遍历原始数组

#include<stdio.h>
#include<string.h>
char a[150] = " qwertyuiopasdfghjkl;zxcvbnm,./";
int main()
{
char c;
char v[150];
while(~scanf("%c%*c", &c))
{
scanf("%s%*c", v);
for(int i = 0; v[i]; i++)
{
for(int j = 1; a[j]; j++)
{
if(a[j] == v[i])
if(c == 'R')
printf("%c", a[j-1]);
else if(c == 'L')
printf("%c", a[j+1]);
}
}
printf("n");
}
return 0;
}

B：二分查找下界，但其实low_bound函数直接可以完成

#include<stdio.h>
#include<string.h>
#define N 100005
int a[N];
int main()
{
int n, m;
while(~scanf("%d", &n))
{
int x;
a[0] = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d", &x);
a[i] = a[i-1] + x;
}
scanf("%d", &m);
while(m--)
{
scanf("%d", &x);
int l = 1, r = n;
while(l < r)
{
int mid = (l + r) / 2;
if(x <= a[mid])
r = mid;
else l = mid + 1;
}
printf("%dn", l);
}
}
return 0;
}

#include <cstdio>
#include <algorithm>
using namespace std;
const int kMaxN = 100000;
int n, s[kMaxN + 1];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++ i) {
scanf("%d", &s[i]);
s[i] += s[i - 1];
}
int m;
scanf("%d", &m);
for (int x; m --; ) {
scanf("%d", &x);
int p = lower_bound(s + 1, s + n + 1, x) - s;
printf("%dn", p);
}
return 0;
}

一共是4*4*4*4种情况，注意的是下标从1开始的话要减回去一个1.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define rep(i, n) for(int i = 1; i <= n; i++)
typedef long long LL;
struct node{
int x, y;
void input()
{
scanf("%d%d", &x, &y);
}
void show()
{
printf("x: %d y: %dn", x, y);
}
}p[10][10], home[5];
LL d[10];
LL dist ( const node& a , const node& b ) {
LL x = a.x - b.x ;
LL y = a.y - b.y ;
return x * x + y * y ;
}
LL solve()
{
LL ans = 999;
rep(i, 4)
rep(j, 4)
rep(k, 4)
rep(o, 4)
{
d[0] = dist ( p[1][i] , p[2][j] ) ;
d[1] = dist ( p[2][j] , p[3][k] ) ;
d[2] = dist ( p[3][k] , p[4][o] ) ;
d[3] = dist ( p[4][o] , p[1][i] ) ;
d[4] = dist ( p[1][i] , p[3][k] ) ;
d[5] = dist ( p[2][j] , p[4][o] ) ;
sort ( d , d + 6 ) ;
if ( d[0] == 0 ) continue ;
else if ( d[0] == d[1] && d[1] == d[2] && d[2] == d[3] && d[4] == 2 * d[0] && d[4] == d[5] )
{
ans = ans < i + j + k + o - 4? ans : i + j + k + o - 4;
}
}
return ans == 999 ? -1: ans;
}
int main()
{
int n;
while(~scanf("%d", &n))
while(n--)
{
rep(i, 4)
{
p[i][1].input();
home[i].input();
p[i][4].x = home[i].x - home[i].y + p[i][1].y ;
p[i][4].y = home[i].y + home[i].x - p[i][1].x ;
p[i][3].x = home[i].x + home[i].x - p[i][1].x ;
p[i][3].y = home[i].y + home[i].y - p[i][1].y ;
p[i][2].x = home[i].x + home[i].x - p[i][4].x ;
p[i][2].y = home[i].y + home[i].y - p[i][4].y ;
}
printf("%dn", solve());
}
return 0;
}

最后

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