A1065 A+B and C (64bit) (20)(20 分) Given three integers A, B and C in [-2^63^, 2^63^], you are supposed to tell whether A+B > C. Input Specification: The first line of the input gives the positive number of test cases, T (<=10). Then T test cases fo…

Given three integers A, B and C in [-263, 263], you are supposed to tell whether A+B > C. Input Specification: The first line of the input gives the positive number of test cases, T (<=10). Then T test cases follow, each consists of a single line co…

AC代码 #include <cstdio> int main() { #ifdef ONLINE_JUDGE #else freopen("1.txt", "r", stdin); #endif // ONLINE_JUDGE char str[2][10] = {"false", "true"}; int n, tcase = 1; scanf("%d", &n); for(int…

判断两数相加是否大于第三数,大于输出true,否则输出false(相等也是false) 1 需要注意数字溢出的问题: 2 先判断溢出,因为在a,b都是负数最小值的情况下,相加直接是正数,在c较小的时候会导致结果出错. #include<cstdio> #include<iostream> using namespace std; int main(){ long long a,b,c; long long sum; int n,res; scanf("%d",&…

Given three integers A, B and C in [−], you are supposed to tell whether A+B>C. Input Specification: The first line of the input gives the positive number of test cases, T (≤). Then T test cases follow, each consists of a single line containing three…

B1018. 锤子剪刀布 (20) Discription: 大家应该都会玩"锤子剪刀布"的游戏:两人同时给出手势,胜负规则如图所示: 现给出两人的交锋记录,请统计双方的胜.平.负次数,并且给出双方分别出什么手势的胜算最大. Input: 输入第1行给出正整数N(<=105),即双方交锋的次数.随后N行,每行给出一次交锋的信息,即甲.乙双方同时给出的的手势.C代表"锤子".J代表"剪刀".B代表"布",第1个字母代表甲方…

1065 A+B and C (64bit) (20 分) Given three integers A, B and C in [−2^​63​​,2​^63​​], you are supposed to tell whether A+B>C. Input Specification: The first line of the input gives the positive number of test cases, T (≤10). Then T test cases follow,…

题目AC汇总 甲级AC PAT A1001 A+B Format (20 分) PAT A1002 A+B for Polynomials(25) PAT A1005 Spell It Right (20) PAT A1006 Sign In and Sign Out (25) PAT A1009 Product of Polynomials(25) PAT A1011 World Cup Betting(20) PAT A1012 Best Rank(25) PAT A1016 Phone B…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…