重庆移动网站制作营销策略有哪些方法

1. 需求分析

将运动分解为鸟的垂直运动和杆的左右运动。

2. 概要设计

2.1 鸟运动部分

2.2 杆的运动

3. 代码实现

#include <stdio.h>
#include <ncurses.h>#include <stdlib.h>
#include <time.h>int vx = 0;
int vy = 1;int bird_r;
int bird_c;int rows;
int cols;int bird_init_x = 5;
int bird_init_pole_gap = 6;int last_fly_time = 1;
int up_time = 3;int pole_width = 3;
int pole_bt_gap = 6;
int pole_in_gap = 6;int pass_pole_nums = 0;
int pole_nums;struct _pole_node {int pole_lb_x;int pole_gap_y;
};typedef struct _pole_node pole_node;pole_node pn[100];
int pole_nxt;enum HIT_STATUS {HIT_NORMAL,HIT_GROUND,HIT_POLE,
};enum GAME_STATUS {GAME_NORMAL,GAME_QUIT,
};void reset_vy()
{vy = 1;
}void set_pole_nums( )
{int tot = cols - bird_init_x - 2;tot -= bird_init_pole_gap;pole_nums = tot /( pole_width + pole_bt_gap );}
void gen_next_pole(pole_node *prev, pole_node *cur )
{if ( !prev || !cur )return;cur->pole_lb_x = prev->pole_lb_x + pole_bt_gap + pole_width;int prev_y = prev->pole_gap_y;int ub = prev_y - pole_bt_gap - pole_in_gap + 1;ub = ub < pole_width + 1 ? pole_width + 1: ub;int lb = prev_y + pole_bt_gap + pole_in_gap - 1;lb = lb > rows - 1 - pole_width  - pole_in_gap ? rows - 1 - pole_in_gap - pole_width : lb;cur->pole_gap_y = ub + rand() % (lb - ub + 1);           }void init_pole_bound()
{// pole_fp = 0;
//    pole_bp = pole_nums - 1;pn[0].pole_lb_x = bird_init_x + bird_init_pole_gap + 1;pn[0].pole_gap_y = ( bird_r - bird_init_pole_gap) + (rand()%(2 * bird_init_pole_gap)); int prev_y;for ( int i = 1; i < pole_nums; ++i) {gen_next_pole( pn + i - 1, pn + i );/*pn[i].pole_lb_x = pn[ i - 1].pole_lb_x  + pole_bt_gap + pole_width;prev_y = pn[ i - 1 ].pole_gap_y;int ub = prev_y - pole_bt_gap - pole_in_gap + 1;ub = ub < pole_width + 1 ? pole_width + 1: ub;int lb = prev_y + pole_bt_gap + pole_in_gap - 1;lb = lb > rows - 1 - pole_width  - pole_in_gap ? rows - 1 - pole_in_gap - pole_width : lb;pn[i].pole_gap_y = ub + rand() % (lb - ub + 1);           */ }
}void bird_fly( )
{vy = -1;up_time = last_fly_time;
}int check_hit( )
{if ( bird_r - 1 == rows )return HIT_GROUND;int nx = pn[pole_nxt].pole_lb_x;int ny = pn[pole_nxt].pole_gap_y;if ( bird_c >= nx && bird_c < nx + pole_width) {if ( bird_r < ny || bird_r >= ny + pole_in_gap )return HIT_POLE;}return HIT_NORMAL;
}void pole_move( )
{for ( int i = 0; i < pole_nums; ++i) {pn[i].pole_lb_x--;int pole_rb = pn[i].pole_lb_x + pole_width - 1;if ( pole_rb < 1) {int idx = ( i - 1 + pole_nums ) % pole_nums;gen_next_pole( pn + idx, pn + i);}}if ( pn[pole_nxt].pole_lb_x + pole_width == bird_c ) {pole_nxt = ( pole_nxt + 1 ) % pole_nums;pass_pole_nums++;}}
void bird_move( )
{bird_r += vy;bird_c += vx; if ( up_time) {up_time--;if ( !up_time ) {vy = 1;}}if ( bird_r < 1)bird_r = 1;
}void draw_wall(  )
{box(stdscr, '#', '#');}
void draw_pole() 
{for ( int j = 0; j < pole_nums; ++j ) {for ( int i = 1; i < rows - 1; ++i) {if ( i >= pn[j].pole_gap_y && i < pn[j].pole_gap_y + pole_in_gap)continue;for ( int k = 0; k < pole_width; ++k) {int curx = pn[j].pole_lb_x + k;int cury = i;if ( curx > 0 && curx < cols - 1 && cury > 0 && cury < rows - 1)mvaddch(cury, curx, '*');}}	   }}void draw_bird( )
{mvaddch( bird_r, bird_c, '@');
}
void draw_tips( )
{int mid_r = LINES / 2;int bg_c = COLS - 20;mvprintw(mid_r, bg_c,"scores: %d", pass_pole_nums );mvprintw(mid_r + 1, bg_c, "Q: quit");}
void draw_frame( )
{clear();draw_wall();draw_bird();draw_pole();draw_tips();refresh();
}void get_rows_cols(int *prows, int *pcols)
{if ( !prows || !pcols)return;*prows = LINES; *pcols = COLS - 20;}void init_ncurse_settings( )
{initscr();noecho(  );timeout( 0 );curs_set( 0 );
}int process_input( int ch )
{switch ( ch ){case 'Q':case 'q':return GAME_QUIT;case 'j':case 'J':bird_fly();break;		}return GAME_NORMAL;
}int main( int argc, char *argv[])
{srand( (unsigned)time(NULL));init_ncurse_settings();get_rows_cols( &rows, &cols );set_pole_nums();bird_r = ( rows - 2) / 2;bird_c = bird_init_x;int ch;init_pole_bound();while ( 1 ) {usleep( 300000 );if ( check_hit() )break;ch = getch();int ret = process_input( ch );if ( ret == GAME_QUIT )break;bird_move();pole_move();draw_frame();}endwin();return 0;
}

4. 效果图