https://pintia.cn/problem-sets/994805342720868352/problems/994805406352654336
Given three integers A, B and C in [−2^63,2^63], you are supposed to tell whether A+B>C我的测试数据如下： 8 -9223372036854775807 -9223372036854775808 3 9223372036854775807 9223372036854775808 -100000000000000 -9223372036854775807 -9223372036854775808 -3 9223372036854775807 9223372036854775808 3 -9223372036854775807 9223372036854775808 -3 -9223372036854775807 9223372036854775808 3 9223372036854775807 -9223372036854775808 -3 9223372036854775807 -9223372036854775808 3
#include<stdio.h>
#include<string.h>
//1065 A+B and C (64bit)
int real_result(char str1[],char str2[],char str3[]){
// str表示a+b 的结果
int i = 0,j = 0;
char str[30];
int carry_flags = 0;
for( j = 0; str1[j] != '\0' && str2[j] != '\0';j++){
int tmpSum = (str1[j] - '0') + (str2[j] - '0');
if(carry_flags) tmpSum++;
if(tmpSum > 9) carry_flags = 1;
else carry_flags = 0;
if(carry_flags){
str[j] = tmpSum % 10 + '0';
}else{
str[j] = tmpSum + '0';
}
}
str[j] = '\0';
//
for(; str1[j] != '\0'; j++){
int tmpSum = str1[j] - '0';
if(carry_flags) tmpSum++;
if(tmpSum > 9) carry_flags = 1;
else carry_flags = 0;
if(carry_flags) str[j] = (tmpSum % 10) + '0';
else str[j] = tmpSum + '0';
}
for(; str2[j] != '\0'; j++){
int tmpSum = str2[j] - '0';
if(carry_flags) tmpSum++;
if(tmpSum > 9) carry_flags = 1;
else carry_flags = 0;
if(carry_flags) (tmpSum % 10) + '0';
else str[j] = tmpSum + '0';
}
if(carry_flags){
str[j] = '1';
str[++j] = '\0';
}else{
str[j] = '\0';
}
//compare two strings
int len_sum = strlen(str),len_c = strlen(str3);
if(len_sum > len_c) return 1;
else if(len_sum < len_c) return 0;
else{ // =
for(i = len_sum - 1; i >= 0;i--){
if(str[i] > str3[i]) return 1;
else if(str[i] == str3[i]) continue;
else return 0;
}
return 2;
}
}
int two_result(char str_a[],char str_b[],char str_c[]){
int flag1 = (str_a[0] == '-')? 1 : 0;
int flag2 = (str_b[0] == '-')? 1 : 0;
int flag3 = (str_c[0] == '-')? 1 : 0;
// 反转为数字字符数组,不包括符号'-'
char new_a[30],new_b[30],new_c[30];
int i = 0,j = 0;
for(i = strlen(str_a) -1; i > 0;i--){
new_a[j++] = str_a[i];
}
if(flag1) // a是负数
new_a[j] = '\0';
else{
new_a[j] = str_a[0];
new_a[++j] = '\0';
}
j = 0;
for(i = strlen(str_b) -1; i > 0;i--){
new_b[j++] = str_b[i];
}
if(flag2) // b is negative
new_b[j] = '\0';
else{
new_b[j] = str_b[0];
new_b[++j] = '\0';
}
j = 0;
for(i = strlen(str_c) - 1; i > 0;i--){
new_c[j++] = str_c[i];
}
if(flag3) // b is negative
new_c[j] = '\0';
else{
new_c[j] = str_c[0];
new_c[++j] = '\0';
}
if(flag1 == 1 && flag2 == 1 && flag3 == 0){ // - - +
return 0;
}
else if(flag1 ==0 && flag2 == 0 && flag3 ==1 ){ // + + -
return 1;
}
else if(flag1 == 1 && flag2 ==1 && flag3 ==1 ){ // - - -
int tmpResult = real_result(new_a,new_b,new_c);
if(tmpResult == 2 || tmpResult == 1) return 0;
else return 1;
}
else if(flag1 == 0 && flag2 ==0 && flag3 == 0 ){ // + + +
int tmpResult = real_result(new_a,new_b,new_c);
if(tmpResult == 2 || tmpResult == 0) return 0;
else return 1;
}
else if(flag1 == 1 && flag2 ==0){
if(flag3 == 1){
int tmpResult = real_result(new_b,new_c,new_a);
if(tmpResult == 2 || tmpResult == 0) return 0;
else return 1;
}else{
int tmpResult = real_result(new_c,new_a,new_b);
if(tmpResult == 2 || tmpResult == 1) return 0;
else return 1;
}
}else if(flag1 == 0 && flag2 == 1){
if(flag3 == 1){
int tmpResult = real_result(new_a,new_c,new_b);
if(tmpResult == 2 || tmpResult == 0) return 0;
else return 1;
}else{
int tmpResult = real_result(new_c,new_b,new_a);
if(tmpResult == 2 || tmpResult == 1) return 0;
else return 1;
}
}
return 0;
}
int main(){
char str1[30],str2[30],str3[30];
int n;
scanf("%d",&n);
for(int i =1; i <= n; i++){
scanf("%s%s%s",str1,str2,str3);
int result = two_result(str1,str2,str3);
if(result == 1)
printf("Case #%d: true\n",i);
else
printf("Case #%d: false\n",i);
}
return 0;
}
重点是：
1、将不等式恒等变形，不拘泥与a+b>c这个表达式。当b为负数是可以是a>c+(-b)，等等。
2、情况要考虑完整，数的符号要识别正确，负数，和非负数；
3、real_result(char str1[],char str2[],char str3[]) 来比较结果大小。考虑进位；
我的测试数据如下：
8 -9223372036854775807 -9223372036854775808 3 Case #1: false 9223372036854775807 9223372036854775808 -100000000000000 Case #2: true -9223372036854775807 -9223372036854775808 -3 Case #3: false 9223372036854775807 9223372036854775808 3 Case #4: true -9223372036854775807 9223372036854775808 -3 Case #5: true -9223372036854775807 9223372036854775808 3 Case #6: false 9223372036854775807 -9223372036854775808 -3 Case #7: true 9223372036854775807 -9223372036854775808 3 Case #8: false
来源：https://www.cnblogs.com/hiwjw/p/12654389.html