Technocup 2017 - Elimination Round 1 C 交互

题目传送门：http://codeforces.com/problemset/problem/727/C

题意：一道交互题，有n个数，你可以询问n次任意两个数的和（这两个数不可以相等），然后猜出n个数分别是多少

思路：之前从没做过交互题，其实很简单。一问一答，先问sum1,2 ,sum2,3,sum1,3算出来a1,a2,a3然后接着问sum3,4,sum4,5........分别算出来a4,a5...就行了

代码如下：

#include <iostream>#include <algorithm>#include <cstring>#include <stdio.h>#include <string>#include <cmath>#include <queue>#include <set>#include <map>#include <stack>#include <bitset>#include <cstdlib>using namespace std;#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   ll            long long#define   ull           unsigned long long#define   mem(n,v)      memset(n,v,sizeof(n))#define   MAX           5005#define   MAXN          10005#define   PI            3.1415926#define   E             2.718281828459#define   opnin         freopen("input.txt","r",stdin)#define   opnout        freopen("output.txt","w",stdout)#define   clsin         fclose(stdin)#define   clsout        fclose(stdout)#define   haha1         cout << "haha1"<< endl#define   haha2         cout << "haha2"<< endl#define   haha3         cout << "haha3"<< endlconst int    INF    =   0x3f3f3f3f;const ll     INFF   =   0x3f3f3f3f3f3f3f3f;const double pi     =   3.141592653589793;const double inf    =   1e18;const double eps    =   1e-8;const ll     mod    =   1e18;const ull    mx     =   133333331;/**************************************************************************/int main(){    int sum[MAX];//    opnin;//    opnout;    int ans[MAX];    mem(ans,0);    int n;    scanf("%d",&n);    printf("? 1 2\n");    fflush(stdout);    scanf("%d",&sum[0]);    printf("? 2 3\n");    fflush(stdout);    scanf("%d",&sum[1]);    printf("? 1 3\n");    fflush(stdout);    scanf("%d",&sum[2]);    int sum_123 = sum[0] + sum[1] + sum[2];    sum_123 /= 2;    ans[2] = sum_123 - sum[0];    ans[0] = sum_123 - sum[1];    ans[1] = sum_123 - sum[2];    for(int i=3;i<n;i++){        int s;        printf("? %d %d\n",i,i+1);        fflush(stdout);        scanf("%d",&s);        ans[i] = s - ans[i-1];    }    printf("!");    for(int i=0;i<n;i++){        printf(" %d",ans[i]);    }//    clsin;//    clsout;}