简单题目，先把时间都归到整年，然后再计算。同时为了防止减法出现xx月00日的情况，需要将d先多增加1，再恢复回来。
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 5 int sy, sm, sd; 6 int ay, am, ad; 7 8 int map[2][13] = { 9 {0,31,28,31,30,31,30,31,31,30,31,30,31}, 10 {0,31,29,31,30,31,30,31,31,30,31,30,31}, 11 }; 12 13 int isLeap(int n) { 14 if ( (n%4==0 && n%100!=0) || (n%400 == 0) ) 15 return 366; 16 else 17 return 365; 18 } 19 20 void subDate(int d) { 21 int i, j, k; 22 23 sy = 2013; 24 d += 283; 25 while (d > isLeap(sy)) { 26 d -= isLeap(sy); 27 --sy; 28 } 29 i = isLeap(sy)==366 ? 1:0; 30 j = 12; 31 while (d > map[i][j]) { 32 d -= map[i][j]; 33 --j; 34 } 35 sm = j; 36 sd = map[i][j] - d + 1; 37 } 38 39 void addDate(int d) { 40 int i, j, k; 41 42 ay = 2013; 43 d += 83; 44 while (d > isLeap(ay)) { 45 d -= isLeap(ay); 46 ++ay; 47 } 48 i = isLeap(ay)==366 ? 1:0; 49 j = 1; 50 while (d > map[i][j]) { 51 d -= map[i][j]; 52 ++j; 53 } 54 am = j; 55 ad = d; 56 } 57 58 int main() { 59 int t, d; 60 61 #ifndef ONLINE_JUDGE 62 freopen("data.in", "r", stdin); 63 //freopen("data.in", "r", stdout); 64 #endif 65 66 scanf("%d", &t); 67 while (t--) { 68 scanf("%d", &d); 69 subDate(d); 70 addDate(d); 71 printf("%04d/%02d/%02d %04d/%02d/%02d ", ay,am,ad, sy,sm,sd); 72 } 73 74 return 0; 75 }