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求4个C语言程序 每个100行左右 大一水平的

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解决时间 2021-01-06 22:35
求4个C语言程序 每个100行左右 大一水平的
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//选手平分,n个选手,m个评委,选手的分数要去掉一个最高分,去掉一个最低分,剩下的取平均分; //裁判的误差是每次给出的分数与选手最后得分的差数之和取平均. #include #define n 5 #define m 4 void main() { float score[n][m], ave_score[n], judge_flag[m][n], ave_juage_flag[m], max,min; int i,j,max_flag=0,min_flag=0; strute player //用来最后给选手排名 { float score; //选手所得分说 int position; //选手第几个出场 }player[n]; for(i=0;iscore[i][j]) { min = score[i][j]; min_flag = j; } } if(max_flag == min_flag) //选手的分数都一样,就去掉第一和第二个裁判给的分 { score[i][0] = 0; score[i][1] = 0; } else { score[i][max_flag] = 0; score[i][min_flag] = 0; } ave_score[i]={0} for(j=0; jscore = max; player[i]->position = max_flag; ave_score[max_flag] = -1; } for(i=0; iscore, player[i]->position);//输出选手排名,分数,及第几位出场 } for(i=0; i
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  • 1楼网友:轻雾山林
  • 2021-01-05 23:42
//制作万年历:以前写的不知道对你有没有用 #include "stdio.h" #include "string.h" #include "stdlib.h" void main(int argc, char** argv) //输入两个参数就打印某年日历,输入三个参数就打印某月,输入某日就打印某日是星期几 { int yue[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}, n, k=0, i, j, y, x, z; char year[12][4] = {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"}, week[7][4] = {"Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun"}; if ( argc < 2 ) { printf( " Parameter is error.\n"); return; } if ( argc >= 2 ) { i = strlen ( argv[1] ); for ( j = 0; j < i; j++ ) //求年数 { y = y*10 + ( argv[1][j] - '0' ); } n = y % 400; n = ( n + (int)( n/4 ) - (int)( n/100 ) )%7; if ( (y%4 == 0) && (y%100 != 0)) //考虑润年情况 { yue[1] = 29; n = (n + 5)%7; } else if ( y%400 == 0 ) { yue[1] = 29; n = (n + 5)%7; } else { n = (n + 6)%7; } if ( argc >= 3 ) { i = strlen ( argv[2] ); for ( j = 0; j < i; j++ ) { x = x*10 + ( argv[2][j] - '0' ); } for ( j = 0; j < x-1; j++ ) { n = n + yue[j]; } n = n % 7; if ( argc >= 4) { i = strlen ( argv[3] ); for ( j = 0; j < i; j++ ) { z = z*10 + ( argv[3][j] - '0' ); } n = ( n + z - 1 )%7; } } } if ( argc == 2 ) { printf ( "%12d", y); for(i = 0; i < 7; i++) { for( j = 0; j < 9; j++) //控制空格数 { printf ( " "); } printf ( "%s", week[i] ); } if ( n == 0 ) { n = 7; } k = n; for ( i = 0; i < 12; i++) { printf("\n"); for( j = 0; j < 9; j++) //控制空格数 { printf ( " "); } printf ( "%s\n", year[i] ); for ( j = 0; j < 12*(k+1); j++ ) //控制空格数 { printf(" "); } for ( j = 1; j <= yue[i]; j++) { printf ( "%12d", j ); k++; if( k%7 == 0) { printf("\n"); for( n = 0; n < 12; n++) //控制空格数 { printf ( " "); } k = 0; } } } } if ( argc == 3) { printf( "%6d%6d\n", y,x); for( j = 0; j < 12; j++) { printf ( " "); } //printf ( " " ); for(i = 0; i < 7; i++) { for( j = 0; j < 3; j++) { printf ( " "); } printf ( "%s", week[i] ); } printf("\n"); if ( n == 0 ) { n = 7; } for( j = 0; j < 13; j++) { printf ( " "); } for ( j = 0; j < 6*n; j++ ) { printf(" "); } k = n; for ( j = 1; j <= yue[x-1]; j++) { printf ( "%6d", j ); k++; if( k%7 == 0) { printf("\n "); for ( i = 0; i < 12; i++ ) { printf(" "); } k = 0; } } } if ( argc >=4 ) { printf("%6d%6d%6d is %s\n", y, x, z, week[n]); } }
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