突然发现网教还能上，就打了打酱油，先贴前几题，没抄题目，只有源程序，适合软院的学弟学妹们参考，后续更不更看心情和水平罢……
都写在一个文件里了，去掉对应的注释，添加需要的头文件（例如string.h、math.h）即可提交。
文章末尾陆续更新……
已有题目：球体问题，修剪草坪，扫雷，合并果子，传送带，贪婪的你，四则运算之加减法，一夜白发《千字文》，自由落体，倒数问题，小浣熊干脆面
贪婪的你，此题要点在于，截止日期最长的若干道题中，分数最高的那道必须放在最后一天做，再向前倒推，每天选择没有截止且没有完成的题目中分数最高的，直到回到第一天，完成一道d>=1且分数最高的题目。
#include <stdio.h> #include <stdlib.h> //6、贪婪的你 #define MAXDAY 10001 typedef struct { int p, d; }P; P item[10001]; int last=0; int cmp(const void * a, const void * b) { return (*(P *)b).d - (*(P *)a).d; } int findMax(int i) { int index = -1, j, maxp = 0; for(j=0; j<last, item[j].d >= i; j++) { if (item[j].d == MAXDAY || item[j].d < i) continue; if (item[j].p > maxp) { maxp = item[j].p; index = j; } } if (index>=0) item[index].d = MAXDAY;//该题目被选中 return index; } int getScore(int n) { int index = 0; if (n==1) return item[findMax(1)].p;//寻找d>=1中分数最高且没做的题 index = findMax(n); if (index != -1) return getScore(n-1) + item[index].p; else return getScore(n-1); } int main() { int i, n; scanf("%d",&n); for (i=0; i<n; i++) { scanf("%d",&item[i].p); } for (i=0; i<n; i++) { scanf("%d",&item[i].d); last = last>item[i].d ? last : item[i].d;//记录最大期限 } qsort(item, n, sizeof(P), cmp);//快排不稳定，令序列按截止日期降序排列 item[n].d = -1;//防止冗余搜索 printf("%d ",getScore(last)); return 0; } /*5、机场传送带 typedef struct{ double s,v; }S; S a[1000005]; int cmp(const void * a, const void * b) { return (*(S *)a).v - (*(S *)b).v; } int main() { int n, m, i, Bi, Ei, Vi; double time = 0, t, x, v, r; scanf("%d",&n); while (n--) { time = 0; scanf("%lf%lf%lf%lf%d",&x,&v,&r,&t,&m); //memset(a, 0, 2*(m+1)*sizeof(a[0][0])); a[0].s = a[0].v = 0; for(i=1; i<=m; i++) { scanf("%d%d%d",&Bi,&Ei,&Vi); a[i].s = Ei - Bi; a[i].v = Vi; x -= Ei - Bi; } a[0].s = x;//视为速度为0的传送带 a[0].v = 0; qsort(a, m+1, sizeof(S), cmp); for (i=0; i<=m; i++) { //printf("len:%d v:%d ",a[i][0],a[i][1]); if (t>0) { x = (r+a[i].v) * t; if (x >= a[i].s) { t -= a[i].s/(r+a[i].v); time += a[i].s/(r+a[i].v); continue; } else { time += t; t = 0; a[i].s -= x; //i--; } } time += a[i].s / (a[i].v+v); } printf("%.6f ",time); } return 0; } */ /*4、合并果子 //直接插入排序 int insertSort(int t, int * a, int n) { int j; if (t == -1) { t = a[0]; for (j=0; j<n-1; j++) { a[j] = a[j+1]; if (t<a[j])//insert t before a[j] { a[j] = t; return 0; } } a[j] = t; } else { for (j=n-1; j>0; j--) { a[j] = a[j-1]; if (t>a[j])//insert t before a[j] { a[j] = t; return 0; } } a[0] = t; } return 0; } int main() { int n = 0, i, tmp, a[10000]={0}, *p, cost = 0; scanf("%d",&n); p = a; for (i=1; i<=n; i++) { scanf("%d",&tmp); insertSort(tmp, a, i); } for (i=0; i<n-1; i++) { p[1] += p[0]; cost += p[1]; insertSort(-1, ++p, n-i-1); } printf("%d ",cost); return 0; } */ /*3、扫雷 int main() { int n, m, i, j, k, l, No=1; char a[102][102];//, short * b[102]; while (1) { scanf("%d%d",&n,&m); if (n*m == 0) break; getchar(); //memset(a, '0', sizeof(a[0][0])*(m+2)*(n+2)); memset(a, '0', sizeof(a[0][0])*10404); for (i=1; i<=n; i++) { for (j=1; j<=m; j++) { scanf("%c",&a[i][j]); } getchar(); } for (i=1; i<=n; i++) { for (j=1; j<=m; j++) { if (a[i][j] == '.') { a[i][j] = '0'; for (k=i-1; k<=i+1; k++) { for (l=j-1; l<=j+1; l++) if (a[k][l] == '*') a[i][j] += 1; } } } } if (No>1) printf(" "); printf("Field #%d: ",No++); for (i=1; i<=n; i++) { for (j=1; j<=m; j++) { printf("%c",a[i][j]); } printf(" "); } } return 0; } */ /*2、修剪草坪 int main() { int i, j, k, l, n, m, * a[100], flag=0; scanf("%d%d",&n,&m); for (i=0; i<n; i++) { a[i] = (int *)malloc(m*sizeof(int)); for (j=0; j<m; j++) { scanf("%d",&a[i][j]); } } for (i=0; i<n; i++) { for (j=0; j<m; j++) { flag = 0; for (k=0; k<n; k++) { if (a[k][j] > a[i][j]) { flag = 1; break; } } for (l=0; l<m; l++) { if (a[i][l] > a[i][j] && flag == 1) { puts("NO"); goto END; break; } } } } puts("YES"); END: for (i=0; i<n; i++) { free(a[i]); } return 0; }*/ /* 1、球体问题 #define PI 2*acos(0.0) int main() { int n=1, r1, r2, d, w; double p, h, s, d1, d2, V, S, density; //printf("PI = %f ",PI); scanf("%d",&n); while (n--) { scanf("%d%d%d%d%lf", &r1, &r2, &d, &w, &s); d1 = r1 - ((r1*r1 - r2*r2 + d*d)/(2.0*d)); d2 = r2 - ((r2*r2 - r1*r1 + d*d)/(2.0*d)); V= PI * ((3.0*r1-d1)* (d1*d1) + (3.0*r2-d2)*(d2*d2)) /3.0; V = PI * (4.0*((r1*r1*r1) + (r2*r2*r2))/3.0)- V; p = (r1+r2+d)/2.0; h = 2*sqrt(p*(p-r1)*(p-r2)*(p-d))/d; S = 2*PI * (d1*r1+d2*r2); S = 4*PI*(r1*r1+r2*r2) - S; density = w/S; printf("%.4f %.4f ", V, S); if (density>s) printf("The Paired-Sphere Sinks. "); else printf("The Paired-Sphere Floats. "); } return 0; } */
#include <stdio.h> #include <string.h> /* 千字文 */ /* | Unicode符号范围 | UTF-8编码方式 n | (十六进制) | (二进制) ---+-----------------------+------------------------------------------------------ 1 | 0000 0000 - 0000 007F | 0xxxxxxx 2 | 0000 0080 - 0000 07FF | 110xxxxx 10xxxxxx 3 | 0000 0800 - 0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx 4 | 0001 0000 - 0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx 5 | 0020 0000 - 03FF FFFF | 111110xx 10xxxxxx 10xxxxxx 10xxxxxx 10xxxxxx 6 | 0400 0000 - 7FFF FFFF | 1111110x 10xxxxxx 10xxxxxx 10xxxxxx 10xxxxxx 10xxxxxx 表 1. UTF-8的编码规则 */ unsigned int unicode_to_utf8(unsigned int unicode, char * utf8) { char * p; if (unicode <0x80) { utf8[0] = unicode; return 0; } else if (unicode < 0x800) { utf8[0] = 0xc0 + ((unicode>>6)&0x1f); p = &utf8[1]; } else if (unicode < 0x10000) { utf8[0] = 0xe0 + ((unicode>>12)&0xf); p = &utf8[2]; } else if (unicode < 0x110000) { utf8[0] = 0xf0 + ((unicode>>18)&0x7); p = &utf8[3]; } else if (unicode < 0x4000000) { utf8[0] = 0xf8 + ((unicode>>24)&0x3); p = &utf8[4]; } else { utf8[0] = 0xfc + ((unicode>>30)&0x1); p = &utf8[5]; } do { *p-- = 0x80 + (unicode & 0x3f); unicode >>= 6; } while (p != &utf8[0]); return 0; } unsigned int utf8_to_unicode(unsigned char * utf8, int inLen) { unsigned int tmp = 0; if (inLen == 6) { tmp += (utf8[5] & 0x3f) + ((utf8[4] & 0x3f)<<6) + ((utf8[3] & 0x3f)<<12) + ((utf8[2] & 0x3f)<<18) + ((utf8[1] & 0x3f)<<24); tmp += ((utf8[1] & 0x3C)<<24) + ((utf8[0] & 0xf)<<30); } if (inLen == 5) { tmp += (utf8[4] & 0x3f) + ((utf8[3] & 0x3f)<<6) + ((utf8[2] & 0x3f)<<12) + ((utf8[1] & 0x3f)<<18); tmp += ((utf8[0] & 0xf)<<24); } if (inLen == 4) { tmp += (utf8[3] & 0x3f) + ((utf8[2] & 0x3f)<<6) + ((utf8[1] & 0x3f)<<12); tmp += ((utf8[0] & 0xf)<<18); } else if (inLen == 3) { tmp += (utf8[2] & 0x3f) + ((utf8[1] & 0x3f)<<6); tmp += ((utf8[0] & 0xf)<<12); } else if (inLen == 2) { tmp += (utf8[1] & 0x3f); tmp += ((utf8[0] & 0x1f)<<6); } else { tmp = utf8[0] & 0x7f; } return tmp; } #define BUFFSIZE 5000 #define HASHSIZE 0x10000 int HashTable[HASHSIZE] = {0}; int main() { int inLen = 0, i = 0, flag = 1; unsigned char text[BUFFSIZE] = {0}; char tmp[8] = {0}; unsigned int unicode = 0; while (scanf("%s",text) != EOF)//(size > 0) { inLen = strlen((const char *)text); for (i=0; i<inLen;) { memset(tmp, 0, 8); if ((text[i] & 0xFC) == 0xFC) { unicode = utf8_to_unicode(&text[i], 6); i += 6; } else if ((text[i] & 0xF8) == 0xF8) { unicode = utf8_to_unicode(&text[i], 5); i += 5; } else if ((text[i] & 0xF0) == 0xF0) { unicode = utf8_to_unicode(&text[i], 4); i += 4; } else if ((text[i] & 0xE0) == 0xE0) { unicode = utf8_to_unicode(&text[i], 3); i += 3; } else if ((text[i] & 0xC0) == 0xC0) { unicode = utf8_to_unicode(&text[i], 2); i += 2; } else//ascii { unicode = utf8_to_unicode(&text[i], 1); i++; } HashTable[unicode] ++; } } for (i=128; i<HASHSIZE; i++) { if (HashTable[i]>1) { unicode_to_utf8(i, tmp); printf("%s ",tmp); printf("0x%04x %d ", i, HashTable[i]); memset(tmp, 0, 8); flag = 0; } } if (flag) printf("No repeat! "); //fclose(f); return 0; }
下文中自由落体的代码其实可以优化时间，没必要走n次循环
//自由落体 #include <stdio.h> #include <math.h> #define EDGE 0.00001 int main() {//H，S1，V，L，K，n （l<=H，S1，V，L，K，n <=100000） double h, s1, v, l, k, t1 = 0, t2 = 0, dY1 = 0, dY2 = 0; int n, count = 0; scanf("%lf%lf%lf%lf%lf%d",&h,&s1,&v,&l,&k,&n);//n是小球个数，实际序号为0 - n-1 while (n--)//以每个球为坐标原点，求抛物线轨迹 { t1 = (s1 - n - EDGE) / v; t2 = (s1 - n + l + EDGE) / v; dY1 = 5 * t1 * t1; dY2 = 5 * t2 * t2; if (dY1 >= h || dY2<=(h-k-EDGE)) continue; count ++; } printf("%d ",count); return 0; }
倒数问题写的逻辑很混乱，没什么参考价值，用到了四则运算题目中的减法子函数
#include <stdio.h> #include <string.h> #include <stdlib.h> //倒数问题 void traverse(char * a, int n) { int i = 0, flag = 0; char * tmp = (char *)malloc(sizeof(char)*n); memcpy(tmp, a, n*sizeof(a[0])); while (n--) { if (!flag && tmp[n]=='0') { continue; } else flag = 1; if(flag) a[i++] = tmp[n]; } a[i] = ' '; free(tmp); } void minus_d(char a, char b, char * c, int * carry) { int tmp = 0; tmp = (a-'0') - (b-'0') - *carry; *carry = tmp>=0 ? 0 : 1;//向上一位借位 *c = (tmp + 10)%10 + '0'; } void minus(char *a, char *b, char *c) {//c = a - b int len1, len2, carry=0, i, j, k=0, cmp; len1 = strlen(a); len2 = strlen(b); cmp = strcmp(a, b); if ((len1==len2 && cmp< 0) || len1<len2) { c[0] = '-'; minus(b, a, &c[1]); return; } else if (cmp == 0) { c[0] = '0'; c[1] = ' '; return; } for (i=len1-1,j=len2-1; i>=0 && j>=0; i--,j--) { minus_d(a[i], b[j], &c[k++], &carry); } for (;i>=0; i--) { minus_d(a[i], '0', &c[k++], &carry); } for (;j>=0; j--) { minus_d('0', b[j], &c[k++], &carry); } traverse(c, k); } int equal_zero(char * a) { int i = strlen(a); while (i--) if (a[i] != '0' && a[i] != '.') return 0; return 1; } #define N 75 int main() { int len = 0, i = 1, time = 0, tmp, m; char a[N] = {0}, b[5*N] = {'1'}, c[N] = {0}; scanf("%d", &m); //getchar();//吸收回车符 while(m--) { memset(a, 0, N); memset(b, 0, 5*N); i=1; time = 0; b[0] = '1'; scanf("%s",a); len = strlen(a);//a的位数 //minus(b, a, c); //if (c[0]=='-') // printf("0."); if (strcmp(a,"1") == 0) { printf("1 "); continue; } printf("0."); while (i<=len) { b[i++] = '0'; minus(b, a, c); if (c[0]=='-') { printf("0"); } else if (equal_zero(c)) { break; } } //b[i] = '0'; while (1) { minus(b, a, c); if (c[0] == '-') { tmp = strlen(b); b[tmp] = '0';// b = b * 10 memset(&b[tmp+1], 0, 5*N-tmp-1); printf("%c",time+'0'); time = 0; continue; } memcpy(b,c,len+1); time ++; if (equal_zero(c)) { printf("%d",time); break; } } printf(" "); } return 0; }
四则运算提供的对外接口：void plus(char * a, char * b, char * c) 和void minus(char * a, char * b, char * c)，其余子函数不要直接调用。另外赠送一个乘法函数void times(char * a, char * b, char * c)。
//7、四则运算之加减法 #include <stdio.h> #include <string.h> void traverse(char * a, int n) { int i = 0, flag = 0; char * tmp = (char *)malloc(sizeof(char)*n); memcpy(tmp, a, n*sizeof(a[0])); while (n--) { if (!flag && tmp[n]=='0') { continue; } else flag = 1; if(flag) a[i++] = tmp[n]; } a[i] = ' '; free(tmp); } void plus_d(char a, char b, char * c, int * carry) { int tmp = 0; tmp = (a-'0') + (b-'0') + *carry; *carry = tmp>9 ? 1 : 0; *c = (tmp % 10) + '0'; } void plus(char *a, char *b, char *c) {//按字节运算 int len1, len2, carry=0, i, j, k=0; len1 = strlen(a); len2 = strlen(b); for (i=len1-1,j=len2-1; i>=0 && j>=0; i--,j--) { plus_d(a[i], b[j], &c[k++], &carry); } for (;i>=0; i--) { plus_d(a[i], '0', &c[k++], &carry); } for (;j>=0; j--) { plus_d('0', b[j], &c[k++], &carry); } if (carry) c[k++] = carry+'0'; traverse(c, k); } void minus_d(char a, char b, char * c, int * carry) { int tmp = 0; tmp = (a-'0') - (b-'0') - *carry; *carry = tmp>=0 ? 0 : 1;//向上一位借位 *c = (tmp + 10)%10 + '0'; } void minus(char *a, char *b, char *c) { int len1, len2, carry=0, i, j, k=0, cmp; len1 = strlen(a); len2 = strlen(b); cmp = strcmp(a, b); if ((len1==len2 && cmp< 0) || len1<len2) { c[0] = '-'; minus(b, a, &c[1]); return; } else if (cmp == 0) { c[0] = '0'; c[1] = ' '; return; } for (i=len1-1,j=len2-1; i>=0 && j>=0; i--,j--) { minus_d(a[i], b[j], &c[k++], &carry); } for (;i>=0; i--) { minus_d(a[i], '0', &c[k++], &carry); } for (;j>=0; j--) { minus_d('0', b[j], &c[k++], &carry); } traverse(c, k); } void times(char *a, char *b, char *c) { int len, shift = 0, i, j; char tmp[2001]; len = strlen(b); shift = strlen(a); c[0] = '0'; c[1] = 0; for (i=0; i<len; i++) { for (j=0; j<b[i]-'0'; j++) { plus(c, a, tmp); memcpy(c, tmp, strlen(tmp)); c[strlen(tmp)] = 0; } if (i < len-1) c[shift++] = '0'; c[shift] = 0; } } int main() { char a[1000]; char b[1000]; char c[2001]; char s[2]; scanf("%s",a); getchar(); scanf("%s",s); getchar(); scanf("%s",b); getchar(); // while (scanf("%s %s %s ", a, s, b) == 3) { if (s[0] == '+') { plus(a, b, c); } else if (s[0] == '-') { minus(a, b, c); } else if (s[0] == '*') { times(a, b, c); } printf("%s ", c); // } return 0; }
//小浣熊干脆面
#include <stdio.h> #include <string.h> #define N 1000000 unsigned short L[N] = {0}; int n, m, record = N; char hash[2001] = {0}; int main() { int i, tmp, count = 1, end = 0; scanf("%d%d",&n,&m); end = n-1; memset(L, N, n*sizeof(L[0])); for (i=0; i<n; i++) { scanf("%d",&L[i]); } hash[L[n-1]] = 1; for (i=n-1; i>0; i--) { if (hash[L[i-1]] == 0) count ++; hash[L[i-1]]++; if (L[i-1] == L[end]) { while (hash[L[end]] > 1) { hash[L[end]]--; end --; } } if (count == m) { tmp = end - i +2; record = record < tmp ? record : tmp;
if (record == m) break; } } printf("%d ", record); return 0; }