MrDoge @ 2022-11-29 21:27:01
求助代码在本地没有问题，在洛谷在线ide上测试，只算一组数的时候没有问题，只要输入多组数就只有最后一组是正常的，剩下的每一组的第二个数都会变成一些奇怪的数，这是什么原理。。。
比如在洛谷ide输入
2
a 10 4
10 4
就会输出
10+5=15
8
10+4=14
7
但是本地真的没有问题
(洛谷要比本地多加一个getchar();才能正常输入也不知道为什么。。。）
#include<cstdio>
#include<cmath>
#include<string.h>
using namespace std;
char all[100], ca[10], cb[10];
int x1[100], x2[100], x3[100], x4[100];
char s1[100];
int main(){
int n;
char f;
int i = 0, cnt = 0, na = 0, nb = 0, numa = 0, numb = 0, out;
scanf("%d", &n);
getchar();
getchar();
for(int k = 1; k <= n; k++){
memset(all, '\0', sizeof(all));
memset(ca, '\0', sizeof(ca));
memset(cb, '\0', sizeof(cb));
gets(all);
if(all[0] == 'a' || all[0] == 'b' || all[0] == 'c'){
switch(all[0]){
case 'a':
f = '+';
cnt = 0, numa = 0, numb = 0;
for(i = 2; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa+numb;
out = numa+numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
case 'b':
f = '-';
cnt = 0, numa = 0, numb = 0;
for(i = 2; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa-numb;
out = numa-numb;
if(numa-numb>=0){
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
else{
cnt = na+nb+2+3;
out = -out;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
break;
case 'c':
f = '*';
cnt = 0, numa = 0, numb = 0;
for(i = 2; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa*numb;
out = numa*numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
}
}
else{
switch(f){
case '+':
cnt = 0, numa = 0, numb = 0;
for(i = 0; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
out = numa+numb;
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa+numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
case '-':
cnt = 0, numa = 0, numb = 0;
for(i = 0; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
out = numa-numb;
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa-numb;
if(numa-numb>=0){
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
else{
cnt = na+nb+2+3;
out = -out;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
break;
case '*':
cnt = 0, numa = 0, numb = 0;
for(i = 0; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
out = numa*numb;
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa*numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
}
}
}
for(int k = 1; k <= n; k++){
printf("%d%c%d=%d\n%d\n", x1[k], s1[k], x2[k], x3[k], x4[k]);
}
}by olegekei @ 2022-11-29 21:32:35
@MrDoge 是gets的问题，你可以读入n也用gets，这样就不用再getchar了
by sunzz3183 @ 2022-11-29 21:48:33
不要用 gets，不要用 gets，不要用 gets！
（不知有多少OIer曾栽在gets上
by MrDoge @ 2022-11-29 22:06:50
@olegekei 我用下面的代码输入了n，在洛谷ide上试着输出了一下也能转换成int型，但整个程序直接没有输出了，一个也算不了了。。。（本地还是没问题）
gets(n);
int n1, n2 = 0;
for(n1 = 0; n[n1] !='\0'; n1++);
for(int k = n1-1; k >= 0; k--){
n2 += (n[cnt]-'0')*pow(10,k);
cnt++;
}完整的代码是下面的（好冗长觉得自己好菜。。）感谢大佬帮助
#include<cstdio>
#include<cmath>
#include<string.h>
using namespace std;
char all[100], ca[10], cb[10];
int x1[100], x2[100], x3[100], x4[100];
char s1[100];
char n[10];
int main(){
int i = 0, cnt = 0, na = 0, nb = 0, numa = 0, numb = 0, out;
gets(n);
int n1, n2 = 0;
for(n1 = 0; n[n1] !='\0'; n1++);
for(int k = n1-1; k >= 0; k--){
n2 += (n[cnt]-'0')*pow(10,k);
cnt++;
}
char f;
for(int k = 1; k <= n2; k++){
memset(all, '\0', sizeof(all));
memset(ca, '\0', sizeof(ca));
memset(cb, '\0', sizeof(cb));
gets(all);
if(all[0] == 'a' || all[0] == 'b' || all[0] == 'c'){
switch(all[0]){
case 'a':
f = '+';
cnt = 0, numa = 0, numb = 0;
for(i = 2; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa+numb;
out = numa+numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
case 'b':
f = '-';
cnt = 0, numa = 0, numb = 0;
for(i = 2; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa-numb;
out = numa-numb;
if(numa-numb>=0){
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
else{
cnt = na+nb+2+3;
out = -out;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
break;
case 'c':
f = '*';
cnt = 0, numa = 0, numb = 0;
for(i = 2; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa*numb;
out = numa*numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
}
}
else{
switch(f){
case '+':
cnt = 0, numa = 0, numb = 0;
for(i = 0; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
out = numa+numb;
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa+numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
case '-':
cnt = 0, numa = 0, numb = 0;
for(i = 0; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
out = numa-numb;
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa-numb;
if(numa-numb>=0){
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
else{
cnt = na+nb+2+3;
out = -out;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
}
break;
case '*':
cnt = 0, numa = 0, numb = 0;
for(i = 0; all[i] != ' '; i++){
ca[cnt] = all[i];
cnt++;
}
cnt = 0;
for(i++; all[i] != '\0'; i++){
cb[cnt] = all[i];
cnt++;
}
for(na = 0; ca[na] != '\0'; na++);
na--;
for(nb = 0; cb[nb] != '\0'; nb++);
nb--;
cnt = na;
for(int j = 0; j <= na; j++){
numa += (ca[j]-'0')*pow(10, cnt);
cnt--;
}
cnt = nb;
for(int j = 0; j <= nb; j++){
numb += (cb[j]-'0')*pow(10, cnt);
cnt--;
}
out = numa*numb;
s1[k] = f;
x1[k] = numa;
x2[k] = numb;
x3[k] = numa*numb;
cnt = na+nb+2+2;
while(out){
cnt++;
out /= 10;
}
x4[k] = cnt;
break;
}
}
}
for(int k = 1; k <= n2; k++){
printf("%d%c%d=%d\n%d\n", x1[k], s1[k], x2[k], x3[k], x4[k]);
}
}by MrDoge @ 2022-11-29 22:09:06
@sunzz3183 因为scanf输入字符串不能带空格，只学了gets就一直用了。。出现了两三次这种情况。。大佬能给几个能输入带空格字符串的函数名吗，我自己查一下学一下。。感谢大佬
by InnitTimmer @ 2022-11-29 22:14:02
@MrDoge getline可以考虑，但是有时会出锅
by MrDoge @ 2022-11-29 22:20:00
@InnitTimmer 好的，感谢
by sunzz3183 @ 2022-11-30 06:51:31
@MrDoge getline比较稳定了，用这个基本不出锅
by MrDoge @ 2022-11-30 09:13:16
@sunzz3183 okk
by sunzz3183 @ 2022-11-30 10:47:26
@MrDoge 这个自动忽略换行
cin.get可以读入换行
getchar的数量由评测环境定，有时win转linux或linux转win就要用两个getchar
by MrDoge @ 2022-11-30 13:59:53
@sunzz3183 明白了，感谢